Question

What is the pH of a solution prepared by mixing 50.0 mL of 0.010 mol/L HCl(aq) and 50.0 mL of 0.010 mol/L Ca(OH)2(aq)? Assume the temperature is 25 degrees celcius.

Answer 1

`pH=11.4........`

Explanation:

`Ca(OH)_2(aq) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)`.

`"Moles of HCl"=50.0xx10^-3Lxx0.010*mol*L^-1=5.00xx10^-4*mol`.

`"Moles of "Ca(OH)_2=50.0xx10^-3Lxx0.010*mol*L^-1=5.00xx10^-4*mol`.

Now there are EQUIMOLAR quantities of `Ca(OH)_2` and `HCl`.

Clearly, at the end of the addition, we have a`100*mL` solution that contains, nominally, `Ca(OH)Cl`, i.e. we have added HALF an EQUIV of acid to the base.

And thus we have `2.500xx10^-4*mol` `HO^-` in `100*mL` of solution.

`[HO^-]=2.50xx10^-4*mol*L^-1`.

`pOH=-log_10([HO^-])=-log_10(2.50xx10^-4)=-(-2.60)=2.60`.

But `pH=14-pOH=14-2.60=11.4`.

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