Question

A sample of potassium phthalate, KHC8H4O4, monobasic, weighing 4.0700g is titrated with NaOH solution and back titrated with HCl. NaOH required = 46.40 mL; HCl required = 5.35mL. If each mL of the HCl is equivalent to 0.01600g of Na2O, what volume of H2O or of 6.00N NaOH must be added to 500mL of the above NaOH to bring it to 0.5000N?

Answer 1

2HCl + Na₂O → 2NaCl + H₂O

M(Na₂O) = 62 g/mol

n(Na₂O) = $\frac{0.016}{62}$ = 0.000258 mol (equivalent)

n(HCl) = 2n(Na₂O) = $2 \times 0.000258$ = 0.000516 mol/ml (equivalent)

Total n(HCl) = $5.35 \times 0.000516$ = 0.00276 mol

HCl + NaOH → NaCl + H₂O

n(NaOH) = n(HCl) = 0.00276 mol

Proportion:

46.40 mL – 0.00276 mol

1000 mL – x

x = $\frac{1000 \times 0.00276}{46.40} = 0.0595 \;M$ (molarity of NaOH)

$C_1V_1 = C_2V_2 \\ 0.0595 \times 46.40 = C_2 \times 500 \\ C_2 = 0.005216 \;M$

C(NaOH) = 0.5 N = 0.5 M

C(NaOH) = 6.0 N = 6.0 M

$C_1 = 6 + 0.0055216 = 6.0055216 \;M \\ V_1 = 500 \;mL \\ C_2 = 0.5 \;M \\ 6.0055216 \times 500 = 0.5 \times V_2 \\ V_2 = 6005.52 \;mL = 6.0055 \;L \\ V(H_2O)_{add} = 6.0055 - 0.5 = 5.5055 \;L$

Answer: 5.505 L