Question

2. What is the molarity of NaOH solution if 35.75 mL of this solution is required to neutralize 1.070 g KHP.

3. What is the molarity of HCl solution if 25.0 mL of this solution required 15.0 mL of 0.050 M NaOH for neutralization?

Answer 1

2. HKC₈H₄O₄ + NaOH → NaKC₈H₄O₄ + H₂O

M(HKC₈H₄O₄) = 204.23 g/mol

n(HKC₈H₄O₄) $= \frac{1.070}{204.23} = 0.00524 \; mol$

n(HKC₈H₄O₄) = n(NaOH) = 0.00524 mol

Proportion:

35.75 mL – 0.00524 mol

1000 mL – x

$x = \frac{1000 \times 0.00524}{35.75} = 0.146 \; M$

Answer: 0.146 M NaOH

3. HCl + NaOH → NaCl + H₂O

Proportion 1 (NaOH):

0.05 mol – 1000 mL

n mol – 15.0 mL

n(NaOH) = $\frac{0.05 \times 15.0}{1000} = 0.00075 \;mol$

n(HCl) = n(NaOH) = 0.00075 mol

Proportion 2 (HCl):

25.0 mL – 0.00075 mol

1000 mL – x

$x = \frac{1000 \times 0.00075}{25.0} = 0.03 \;M$

Answer: 0.03 M HCl