In the following problems, I have a mixture consisting of 0.180M nitrous acid, a

Question

In the following problems, I have a mixture consisting of 0.180M nitrous acid, and 0.150M sodium nitrite. Note: pka (nitrous acid)=3.14. a) what is the pH of the nitrous acid/sodium nitrate) mixture? b) I add 0.010 mol of HCl to 1.00 L of the (nitrous acid/sodium nitrite) mixture, without changing the volume of the solution in the process. What is the new pH of the resulting mixture? c) Next, I add 0.010 mol of HCl to 1.00 L of water, without changing its volume. What is the new pH of the resulting mixture in this case? d)Secondly, I add 0.010 mol of NaOH to 1.00 L of the original (nitrous acid/sodium nitrite) mixture, without changing the volume of the solution in the process. What is the new pH of the resulting mixture? e) Finally, I add 0.010 mol of NaOH to 1.00 L of water, without changing its volume. What is the new pH of the resulting mixture in this case?

Explanation / Answer

I have a mixture consisting of 0.180M nitrous acid, and 0.150M sodium nitrite. Note: pka (nitrous acid)=3.14. a) what is the pH of the nitrous acid/sodium nitrate) mixture?

pH=pka+log[sod nitrite]/[nitrous acid]=3.14+log(0.15/0.18)=3.14-0.079=3.061

b) I add 0.010 mol of HCl to 1.00 L of the (nitrous acid/sodium nitrite) mixture, without changing the volume of the solution in the process. What is the new pH of the resulting mixture?

pH=3.14+log(0.15-0.01/0.18+0.01)=3.14+log(0.14/0.19)=3.14-0.133=3.007

c) Next, I add 0.010 mol of HCl to 1.00 L of water, without changing its volume. What is the new pH of the resulting mixture in this case?

initial pH of H2O=7

molarity of HCl=0.01mole/L

pH=-log(0.01)=2

pH change=5units

d)Secondly, I add 0.010 mol of NaOH to 1.00 L of the original (nitrous acid/sodium nitrite) mixture, without changing the volume of the solution in the process. What is the new pH of the resulting mixture?

pH=3.14+log(0.15+0.01/0.18-0.01)=3.14+log(0.16/0.17)=3.14-0.026=3.114

e) Finally, I add 0.010 mol of NaOH to 1.00 L of water, without changing its volume. What is the new pH of the resulting mixture in this case?

initia pah of H2O=7

Molarity of NaOH=0.01mole/1L=0.01M

pOH=-log(0.01)=2

PH=14-2=12

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