Question

What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution

Answer 1

Answer:

The volume required for complete neutralize is 32.29 mL

Explanation:

The computation of the volume required for complete neutralize is shown below:

As we know that, the balanced equation is

$Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O$

Now

The number of moles of $Ba(OH)_2$ = n_1 = 1

And, the number of moles of Hcl = n_2 = 2

Therefore

The equation i.e. to be used to find out the volume is given below:

$\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$

$V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}$

= 32.29 mL

Hence, the volume is 32.29mL