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.

Question

asked Jul 12, 2019 67.3k views

1 Answer

Deathspike · Answer 1 · 2019-07-16T10:39:16+0000

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((1L)/(1000mL))((0.3molHCl)/(1L))

= 0.075 mol HCl

100.mL((1L)/(1000mL))((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 resulting solution = 0.250 L + 0.100 L = 0.350 L

So, the concentration of HCl in the resulting solution =
(0.050mol)/(0.350L)

= 0.143 M

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