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.

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