Question

What is the pH of a solution that results from adding 0.053 mol Ba(OH)₂ (s) to 1.57 L of a buffer comprised of 0.31 M HF and 0.31 M NaF?

its not 3.40

Answer 1

Moles of HF = Molarity*Volume = 0.31M*1.57L = 0.487moles

Moles of NaF = 0.31M*1.57L = 0487moles

Moles of OH^- from Ba(OH)2 = 2*0.053mol = 0.106moles

When Ba(OH)2 is added-

HF + OH^- ---> F^- + H2O

Initial 0.487mol 0 0.487mol

Adding of 0.106moles

Change -0.106 -0.106 +0.106mol

Final 0.381moles 0.593moles

So, [F^-] = 0.593moles

[HF] = 0.381moles

Henderson-Hasslebach equation-

pH = pKa + log[salt/acid]

Ka for HF = 6.6*10^-4

pKa = -log(ka) = -log(6.6*10^-4)

= -log6.6+4

= 3.18

Keeping all the values,

pH = 3.18 + log[0.593/0.381]

= 3.18 + log1.556

pH = 3.37