1.)How to calculate the ph of a solution containing 0.01 moles H2So4 and 0.01 moles KOH in 2 litre water (log 5=0.699)? 2.)ph of 10^-8 molar Ca(OH)2

Asked by  | 3rd Oct, 2013, 06:22: PM

Expert Answer:

(1) 

H2SO4 will dissociate as:

H2SO4   ?  2H+    +   SO42-

0.01 moles of H2SO4 will give 2 X 0.01 = 0.02 moles of H+.

KOH will dissociate as:

KOH ? K+ + OH-

0.01 moles of KOH will give 0.01 moles of OH-.

Out of 0.02 moles of H+ from H2SO4, 0.01 moles of H+ will be neutralized by 0.01 moles of OH- from KOH. The number of moles of OH- remaining in 2 L water will be:

 0.02 - 0.01 = 0.01

Concentration of H+ = 0.01moles / 2L = 0.005 moles/L = 5 x 10-3 moles/L

(2)

Ca(OH)2  ?  Ca2+ + 2 OH-
[OH-] = 2 x 10-8 M
[OH-] by the auto ionization of water = 1 x 10-7
total [OH-] = 1 x 10-7 + 2 x 10-8 = 1.2 x 10-7 M
pOH = - log 1.2 x 10-7 = 6.9
pH = 14 - pOH = 7.1

Answered by Hanisha Vyas | 3rd Oct, 2013, 06:56: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.