How to calculate the pH of this solution? 0.002 mol of solid calcium hydroxide is added in 250 mL solution of hydrochloric acid with the concentration of 0.004 mol / L.

1 Answer
anor277
Jan 31, 2018

Well, first we interrogate the stoichiometric equation....

Explanation:

#Ca(OH)_2(s) + 2HCl(aq) rarr CaCl_2(aq) +2H_2O#..

And then we determine equivalent quantities of (i),

#"Calcium hydroxide"=0.002*mol#...

...and (ii) hydrochloric acid....

#n_"HCl"=250*cancel(mL)xx10^-3*cancel(L*mL^-1)xx4xx10^-3*mol*cancel(L^-1)#

#=0.001*mol#...

...and given that calcium hydroxide is a dibase....we gots #0.003*mol# #HO^-# dissolved in #250*mL# of solution...

#[HO^-]=(0.003*mol)/(0.250*L)=0.012*mol*L^-1# with respect to hydroxide anion...

Now............

#pOH=-log_10[HO^-]=-log_10(0.012)=-(-1.92)=1.92#

But #pH=14-pOH=14-1.92=12.08#

Note that I have assumed (reasonably) that all the calcium hydroxide has gone up into solution...I am not going to consider the aqueous solubility of calcium hydroxide....

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