Here we’ll go over an example where a strong acid is mixed with a strong base, and we calculate the pH of the final mixture.

Strong Acid–Strong Base

Mixture CalculationsExample 1

We’re given that 150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.

And we’re asked to calculate the pH of the final mixture.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

We’ll point out something important here. Strontium hydroxide is a strong base,

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

22Sr 2 HSr OH O

and its formula, Sr(OH)2, has 2 OH’s in it.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

22Sr 2 HSr OH O

So when we write the balanced dissociation equation for Sr(OH)2.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

22

Sr OH Sr 2OH Dissociation Equation

We see that there are 2 moles of OH- for each mole of Sr(OH)2.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

22SrSr OH 2OH

2

2 mol OH

1 mol Sr OH

A conversion factor

So we can use this as a conversion factor

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

22SrSr OH 2OH

2

2 mol OH

1 mol Sr OH

A conversion factor

We’ll start by calculating the initial moles of OH minus added.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OH0.150L 0.0600mol OH

1L 1molSr(mol OH

OH)initial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

We take 0.200 moles of Sr(OH)2 per L

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

2mol OHmol OH 0.150L 0.0600mol OH

1molSr(OH)

0.200molSr(OH)

1Linitial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Multiply it by the conversion factor 2 mol OH- over 1 mole Sr(OH)2.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

2mol0.200molSr(OH)mol

OH

1molSr(OH)OH 0.150L 0.0600mol OH

1Linitial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

And by 0.150 L. We rounded this to 3 significant figures to save room here. The concentration 0.200 mol/L is only 3 significant figures so the answer to this calculation is limited to 3 significant figures.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.0600mol OH

1L 1molSr(0.150L

OH)initial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

When we multiply all three numbers we get 0.0600 mol. So the initial moles of OH minus added is 0.0600 mol. Notice, this is expressed to 3 significant figures, which is consistent with the given data.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L

1L 1molSr(0.0600mol H

OH)Oinitial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Our next step is to calculate the initial moles of H+ added.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

0.100mol HNO 1mol H0.350L 0.0350mol H

1L 1mol Hmol

NOHinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

The H+ comes from the strong acid nitric acid, or HNO3. Each HNO3 releases 1 proton, so we take 0.100 mol HNO3 per L

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

1mol Hmol H 0.350L 0.0350mol H

1mol HNO

0.100mol HNO

1Linitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Times 1 mol of H+ per 1 mol of HNO3

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

1mo0.100mol HNOmol

l H

1mol HNOH 0.350L 0.0350mol H

1Linitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Times 0.350 L. Again we rounded this to 3 significant figures to save room. The number of significant figures in the answer is limited by the 3 significant figures in 0.100 M.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

0.100mol HNO 1mol Hmol H 0.0350mol H

1L 1mol H0.350L

NOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

The answer comes out to 0.0350 mol. So the initial moles of H+ added is 0.0350 moles.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

0.100mol HNO 1mol Hmol H 0.350L

1L 1mol H0.0350mo H

NOlinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Notice that in preserving 3 significant figures, both of these have 4 decimal places

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0. mol OH

1L 1molSr(060

OH)0initial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0. mol H

1L 1mol H03

N5

O0initial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Comparing the initial moles of OH minus with the initial moles of H+, we see that we have more moles of OH minus than of H+, so the OH minus is in excess in this case.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L

1L 1molSr(0.0600mol H

OH)Oinitial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol H 0.0600mol OH 0.0350mol H 0.0250mol OHinitial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Excess

The excess moles of OH minus

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

0.0600mol OH 0.0Excessmol 350mol H 0.0250moO lH OH

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

Is 0.0600 mol OH minus

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L

1L 1molSr(0.0600mol H

OH)Oinitial

0.0600molExcessmol OH 0.0350mol H 0.0250moH l OHO

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

minus 0.0350 mol H+

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0.0600mol O 0.0350mH 0.0250mol Hol H O

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L

1L 1mol H0.0350mo H

NOlinitial

Which equals 0.0250 mol OH minus

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0.0600mol OH 0.0350m 0.0250mol Hol H O

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Notice the numbers we’re subtracting both have 4 decimal places, so our answer must also have 4 decimal places.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0. mol O0600 035H 0. mol H 0. mo0 l O0 H025

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

4 decimal places

4 decimal places

4 decimal places

When expressed to 4 decimal places, this number has 3 significant figures. The zero’s to the left of the 2 are not significant, but the zero after the 5 is.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0 mol0 OH25

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

3 significant figures

Since the hydroxide ion is in excess, we calculate its concentration in the final mixture. It is equal to

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

0.0250mol 0.0250mol

0.0500M0.150L 0.350L 0.500L

OH

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

0.0250 moles

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

Excessmol OH 0.0600mol OH 0.0350m 0.0250mol Hol H O

0.0250mol

OH 0.0500M0.150L 0.350L 0.500

0.0250mol

L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Divided by the total volume of the solution, which is 0.150 L of Strontium hydroxide solution,

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.0600mol OH

1L 1molSr(0.150L

OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.350L 0.500L0.150L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Plus 0.350 L of HNO3 solution.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.350L

0.0250mol 0.0250molOH 0.0500M

0.150L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.0350mol H

1L 1mol H0.350L

NOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

So the concentration of OH minus is 0.0250 moles over 0.500 L

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol

OH 0.0500M0.150L 0.35

0.0250mol

0.5000L L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Which Equals 0.0500 molar

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

0.150L 0.350L 0.500OH 0.0500M

L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Because we have base in excess, we can calculate the pOH

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

log OH log 0.0500 1.30pOH 1 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Which is the negative log of the hydroxide ion concentration

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log 0.0500 1.3log OH 01 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Or the negative log of 0.0500

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

0.150L 0.350L 0.500OH 0.0500M

L

pOH log OH 1.3log 0.05 0100 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Which comes out to 1.301 . None of our data or calculations in this problem have less than 3 significant figures, so the pOH has 3 significant figures or three decimal places.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

1.30pOH log OH log 0. 500 10 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Now we can calculate the pH

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 14.000 pOH 14.000 1.301p 12.699H

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Which is 14 minus the pOH

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.00014.000 1.3p 01 12.H 9O 69

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Or 14 minus 1.301

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

1.30pOH log OH log 0. 500 10 14.000pH 14. 1.3000 pO 01H 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

Which is 12.699. This must has 3 significant figures so its expressed to 3 decimal places.

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 pH 14.000 pOH 14.000 1.301 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

We have now answered the question we set out to answer. The pH of the final mixture is 12.699. This is relatively highly basic which is consistent with the fact that OH- is in excess.

2

2

0.200molSr(OH) 2mol OHmol OH 0.150L 0.0600mol OH

1L 1molSr(OH)initial

0.0250mol 0.0250mol

OH 0.0500M0.150L 0.350L 0.500L

pOH log OH log 0.0500 1.301 14.000 pOH 14.000 1.301pH 12.699

3

3

0.100mol HNO 1mol Hmol H 0.350L 0.0350mol H

1L 1mol HNOinitial

Excessmol OH 0.0600mol OH 0.0350mol H 0.0250mol OH

pH of final mixture

150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3. Calculate the pH of the final mixture.