Question

An antacid tablet weighing 1.30 g was fully neutralized in 42.00 mL (an excess amount) of 0.250 M HCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize?

Answer 1

HCl + NaOH → NaCl + H₂O

C(HCl) = 0.25 mol/L

V(HCl) = 42 ml = 0.042 L

$n(HCl) = 0.25 \times 0.042 = 0.0105 \;mol$

C(NaOH) = 0.1 mol/L

V(NaOH) = 10 ml = 0.01 L

$n(NaOH) = 0.1 \times 0.01 = 0.001 \;mol$

Moles of acid neutralized by the antacid tablet = 0.0105 – 0.001 = 0.0095 mol

Answer: 0.0095 mol