Question

1. How many grams of NaF would have to be added to 2.00L of 0.100M HF to yield a
solution with a pH = 4.00?

2.) 100.0 ml of a solution of a weak, monoprotic acid required 32.00ml of a 0.1500M
KOH for titration to the equivalence point. After adding 24.00ml of KOH solution, the
pH was observed to be 7.97. What is the Ka for the acid.

Answer 1

Ka FOR HF = 6.6*10^-4

pKa = - log (Ka)
= - log(6.6*10^-4)
= 3.180456

use formula for buffer
pH = pKa + log ([NaF]/[HF])
4.00 = 3.180456 + log ([NaF]/[HF])
log ([NaF]/[HF]) = 0.819544
[NaF]/[HF] = 6.6
[NaF] = 0.66
volume , V = 2.0 L
number of mol,
n = Molarity * Volume
= 0.66*2
= 1.32 mol

Molar mass of NaF,
MM = 1*MM(Na) + 1*MM(F)
= 1*22.99 + 1*19.0
= 41.99 g/mol

mass of NaF,
m = number of mol * molar mass
= 1.32 mol * 41.99 g/mol
= 55.43 g

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