[kwl2004]

5.2g of a dibasic acid H₂A was dissolved in 250cm³ of water. In a titration, 20.0cm³ of the acid required 40.0cm³of a solution of NaOH for the reaction.Determine a) the concentration of NaOH in moldm^-3
                           b) concentration of OH^- in gdm^-3.
Given M_r of H₂A = 104

This is what I have done, can anyone please help me check whether it is correct

Workings:

a)
H²A + 2NaOH -> Na₂A + 2H₂0

no. of mol of H₂A in 250cm³ = 5.2/104 = 0.05 mol

no. of mol of H₂A in 20cm³ = (20/250) * 0.05 = 0.004 mol

1 mol of H₂A will react with 2 mol of NaOH

0.004 mol of H₂A will react with 0.008 mol of NaOH

concentration of NaOH = 0.008/ (40 * 10^-3) = 0.04 moldm^-3


b)
NaOH -> Na⁺ + OH^-

0.004 mol of NaOH will react with 0.004 mol of OH^-

concentration of OH^- = 0.04 moldm^-3

M_r of OH = 16.0 + 1.0 = 17.0

concentration of OH^- in gdm^-3 = (0.04 *17.0) = 0.68gdm^-3

[KyleDiLeo]

I wish I wasn't so rusty in chemistry right now and used those units... But from what I can tell it looks like a logical process except that I don't know what moldm^-3 or gdm^-3, but granted that translates into the right units in the end for concentration, it looks correct.