Answer to Question #248139 in Inorganic Chemistry for Aga

Question

Resolve the concentration of 10 mL of 0.1 N hydrochloric acid accidentally added with 90 mL of distilled water.

Answer 1

Proportion:

0.1 mol – 1000 mL

x mol – 10 mL

$x = \frac{0.1 \times 10}{1000}=0.001 \;mol$

We have 0.001 mol of HCl in 10 ml of 0.1 N solution.

At the next step we have the same amount of HCl (0.001 mol) in (10+90=100 mL of solution).

Proportion:

0.001 mol – 100 mL

x mol – 1000 mL

$x = \frac{0.001 \times 1000}{100}=0.01 \;N$

Answer: 0.001 N