What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [Given: `log_(10) (45) = 1.6532`]

Answer

`12.65`

Explanation

Assuming that 1000 mL of 0.1 M NaOH is mixed with 1000 mL of 0.01 M HCl.
`underset(0.1" mol")(NaOH) + underset(0.01" mol")(HCl to) underset(0.01" mol")(NaCl +H_(2)O)`
Here, NaOH is the excess reagent. Hence, the pH of the solution will be greater than 7.0 Excess amount of NaOH = 0.09 mol The volume of the resulting solution `~~2000mL ~~2L`
`:. [OH^(-)]=(0.09)/(2)=0.045` mol/L
`pOH = -log_(10)[OH]`
`= - log_(10) (45 xx 10^(-3))`
`= -[1.6532-3]= 1.3468`
`pH +pOH =14`
`:.pH=14-pOH =14-1.3468`
`=12.6532~~12.65`