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

The volume required for complete neutralize is 32.29 mL

Step-by-step 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:

`(M_1V_1)/(n_1) = (M_2V_2)/(n_2)`

`V_2 = (M_1V_1)/(n_1) * (n_2)/(M_2) \\\\ = (0.2350 * 23.55)/(1) * (2)/(0.3428) \\\\ = (11.0685)/(0.3428)`

= 32.29 mL

Hence, the volume is 32.29mL