Question

What will be the final pH when 5.865 mL of 3.412 M NaOH is added to 0.5000 L of 1.564 x 10-3 M HCl?

Answer 1

Answer: 12.5

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`Molarity=\frac{moles}{\text {Volume in L}}`

moles of
`NaOH=Molarity* {\text {Volume in L}}=3.412* 5.865* 10^(-3)L=0.02moles`

moles of
`HCl=Molarity* {\text {Volume in L}}=1.564* 10^(-3)* 0.5000L=0.782* 10^(-3)moles`

`NaOH+HCl\rightarrow NaCl+H_2O`

According to stoichiometry:

1 mole of
`HCl` require 1 mole of
`NaOH`

Thus
`0.782* 10^(-3)moles` will combine with
`0.782* 10^(-3)moles` of
`NaOH`

Thus
`(0.02-0.782* 10^(-3))moles=0.019moles` of
`NaOH` will be left

Thus Molarity of
`OH^-=(0.019)/(0.56L)=0.03M`

`pOH=-\log[OH^-]`

Putting in the values:

`pOH=-\log[0.03]`

`pOH=1.5`

`pH+pOH=14`

`pH=14-1.5=12.5`

Thus final pH will be 12.5.