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.

1 Answer
Apr 26, 2017

Answer:

#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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