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
[tex]Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O[/tex]
Now
The number of moles of [tex]Ba(OH)_2[/tex] = 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:
[tex]\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}[/tex]
[tex]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}[/tex]
= 32.29 mL
Hence, the volume is 32.29mL
