Question

1.) A buffer solution was prepared by mixing 100. mL of 0.500M HF (K_a = 6.6 $LaTeX: \times$ × 10^-4) and 250. mL of 0.250M NaF. You plan to destroy this buffer by addition of KOH. What would be the minimum amount (in grams) of KOH to destroy this buffer?

a. 5.94 g

b. 3.95 g

c. 2.81 g

d. 7.27 g

e. 4.25 g

Answer 1

Mol of HF present = M(HF)*V(HF)
= 0.500 M * 0.100 L
= 0.0500 mol

So, we can add 0.0500 mol of KOH before buffer is destroyed

Molar mass of KOH,
MM = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol

use:
mass of KOH,
m = number of mol * molar mass
= 5*10^-2 mol * 56.11 g/mol
= 2.81 g
Answer: c