Question

Use the Henderson–Hasselbalch equation to calculate the pH of each solution: Part A a solution that contains 0.625% C 5 H 5 N by mass and 0.820% C 5 H 5 NHCl by mass Part B a solution that is 16.0 g of HF and 24.0 g of NaF in 125 mL of solution

Answer 1

Henderson-Hasselbalch equation is used to find out pH of solution in any chemical or biological system.

It is given by:

where [HA]= molar concentration of undissociated weak acid (proton donor)

[ A^-]= molar concentration of conjugate base or salt

pKa = constant (negative log of acid dissociation constant)

PART A:

• 100gm of solution contains 0.625 gm of C₅H₅N & 0.820gm of C₅H₅NHCl
• molar conc. of weak acid C₅H₅NHCl [HA]= 0.820/ 115.5 = 0.00709 M
• molar onc. of conjugate base C₅H₅N [A^-]= 0.625/ 79.1 = 0.0079 M
• pKa for pyridine = 5.23
• putting all values to the above equation : pH= 5.23+0.045 = 5.28

ANSWER: pH of the solution = 5.28

PART B :

• 125ml of solution contains 16gm of HF & 24gm of NaF
• Molar Conc. of weak acid HF =16/125*20 =0.0064 M
• Molar Conc. of conjugate base NaF =24/125* 42 = 0.0045 M
• pKa value for HF= 3.17
• putting all the values to above equation: pH= 3.17 + (-0.153) = 3.02

ANSWER: pH of the solution= 3.02