What is the pH of a solution that results from adding 0051 m

What is the pH of a solution that results from adding 0.051 mol KOH (s) to 1.75 L of a buffer comprised of 0.387 M HF and 0.387 M NaF? Submit Answer Tries 0/3

Solution

Moles of HF = 0.387 M x 1.75 L = 0.677 moles
Moles of NaF = 0.387 M x 1.75 L = 0.677 moles

When 0.051 moles of KOH is added, it will react with 0.051 moles of HF to produce 0.051 moles of NaF. So, moles of HF will decrease and moles of NaF will increase.

Moles after addition of KOH are
Moles of HF = 0.677 mol - 0.051 mol = 0.626 mol
Moles of NaF = 0.677 mol + 0.051 mol = 0.728 mol

Total volume = 1.75 L

So, the concentration terms are

[HF] = 0.626 mol / 1.75 L = 0.358 M
[NaF] = 0.728 mol / 1.75 L = 0.416 M

pKa of HF= 3.17

Now, using Henderson-hessalbalach equation, we get;

pH = pKa + log { [NaF] / [HF] }
     = 3.17 + log (0.416 /0.358)
     = 3.17 + log (1.16)
     = 3.17 + 0.06
     = 3.23