Question

A 0.1873 g sample of a pure, solid acid, h2x was dissolved in water and titrated with 0.1052 m naoh solution. the balanced equation for the neutralization reaction occurring is h2x(aq) + 2naoh(aq)  na2x(aq) + 2h2o(l) if the molar mass of h2x is 85.00 g/mol, calculate the volume of naoh solution needed in the titration.

Answer 1

According to the balanced equation of the reaction:
H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)

First, we have to get the no. of moles of H2X:
no.of moles of H2X = weight / molar mass
when we have the H2X weight = 0.1873 g & the molar mass H2X = 85 g/mol
So by substitution:
∴ no.of moles of H2X = 0.1873 /85
                                     = 0.0022 mol
-then, we need to get no.of moles of NaOH:
from the balanced equation, we can see that 1 mol H2X = 2 mol NaOH
∴ no.of moles of NaOH = no.of moles of H2X *2
                                       = 0.0022 * 2 = 0.0044 mol
So we can get the volume per litre from this formula:
M (NaOH) = no.of moles NaOH / Volume L
So by substitution:
0.1052 = 0.0044 / Volume L
∴Volume = 0.042 L *1000 = 42 mL