A 1 liter solution contains 0.247 M nitrous acid and 0.329 M sodium nitrite. Addition of 0.271 moles of calcium hydroxide will: (Assume that the volume does not change upon the addition of calcium hydroxide.)

a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity

a. Raise the pH slightly

Explanation:

We know that

Pka of HNO2/KNO2 =3.39

Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles

Moles of NO2-=0.329mol/L×1L=0.329 moles

If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles

Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles

pH=pka+log [base]/[acid]      {henderson -hasselbach equation}

=3.39+log (0.0693/0.0317)=3.39+0.34=3.73

pH=3.73

the answer is:  

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