Question

The titration of 25 mL of 2.0 M NaOH reaches the equivalence or end point when 20 mL of an unknown concentration of HCl has been added. What is the molar concentration of the unknown HCl?

Answer 1

VA 20ml-0.02L
VB 25ml-0.025L
CB 2.0M
CA (Unknown)

Working.
CAVA=CBVB
CA × 0.02 = 2.0 × 0.025
0.02CA = 0.05
0.02CA/0.02 = 0.05/0.02
CA = 2.5
Therefore the concentration of the acid (CA) is 2.5 Molar.

Answer 2

The molar concentration of the unknown HCl is 2.5 M.

Step-by-step explanation:

`HCl+NaOH\rightarrow H_2O+NaCl`

Moles of NaOH in 25 mL of 2.0 M solution = n

`2.0 mol/L=(n)/(0.025 L)`

n = 0.05 moles

According to reaction 1 mol of NaOH reacts with 1 mol of HCl then 0.05 mol of NaOH will react with 0.05 moles of HCl.

Molar concentration of the unknown HCl solution.

[tex}M=\frac{\text{Number of moles of HCl}}{\text{Volume of solution in L}}[/tex]

`M=(0.05 mol)/(0.020 L)=2.5 mol/L`

The molar concentration of the unknown HCl is 2.5 M.