Question

A 250. ML sample of 0.3M HCl is partially neutralized by the addition of 100. ML of 0.25M NaOH.Find the concentration of hydrochloric acid in the resulting solution.

Answer 1

Answer:- 0.143 M

Solution:- HCl and NaOH reacts in 1:1 mol ratio as shown in the below reaction:

$HCl(aq)+NaOH(aq)\rightleftharpoons NaCl(aq)+H_2O(l)$

Let's calculate the initial moles of HCl and the moles of NaOH added to it:

$250.mL(\frac{1L}{1000mL})(\frac{0.3molHCl}{1L})$

= 0.075 mol HCl

$100.mL(\frac{1L}{1000mL})(\frac{0.25molNaOH}{1L})$

= 0.025 mol NaOH

Since they react in 1:1 mol ratio, 0.025 mol of NaOH will react with 0.025 moles of HCl.

Remaining moles of HCl = 0.075 - 0.025 = 0.050

Total volume of the solution = 0.250 L + 0.100 L = 0.350 L

So, the concentration of HCl in the resulting solution = $\frac{0.050mol}{0.350L}$

= 0.143 M

Hence, the concentration of HCl acid in the resulting solution is 0.143 M.