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....

Related questions
Impact of this question
2539 views around the world
You can reuse this answer
Creative Commons License