Question

A is a solution of 0.01molperdm³, HCL.

B is a solution of sodium hydroxide

I. Calculate the average titre value

II. Calculate the concentration of B in mol/dm³

III. Concentration of B in grams per dm³

IV. The number of moles of acid in average titre

V. The number of moles of alkaline in the volume of B pipette .

Answer 1

Reaction between NaOH and HCl can be given as follows $NaOH+ HCl \longrightarrow NaCl+ H_2O$

Molarity of HCl is 0.01 M, mole of HCl react with NaOH = 0.01

(i) Average titrate value $= \frac{V_1+V_2}{2}= \frac{0.01+0.01}{2}=0.01$

(ii) concentration of B = 0.01 molar

(iii) HCl is acid and its molarity is 0.01 so number of moles of acid is 0.01

(iv) Since molarity is same for both then mole of alkalines =0.01