Calculate the pH and pOH of each of the following aqueous solutions of strong acd or base:

Question

(a) 0.010 M HNO3(aq)
(b) ) 1.0 x 10-3M Ba(OH)2(aq)
(c) 10.0 mL of 0.22M KOH(aq) after dilution to 250 mL
(d) 100 mL aqueous solution containing 050 g of Ba(OH)2 (aq)

Answer 1

Here is how you do a and b.
a.
HNO3 is a strong acid; i.e., it ionizes 100%. Therefore, HNO3 ==> H^+ + NO3^- amd H^+ = 0.01
Then pH = -log(H^+) = -log(0.01) = -(-2) = 2

b.
Ba(OH)2 is a strong base and ionizes 100% as in Ba(OH)2 ==> Ba^2+ + 2OH^-
In a above you had only 1 H^+ in HNO3; here you have 2 OH^- in Ba(OH)2; therefore, (OH^-) is twice Ba(OH)2 so (OH^-) = 2*1E-3 = 2E-3
pOH = -log(OH^-) = -log(2E-3) = -(-2.699) = -2.699 which I would round to 2.7. To convert to pH use
pH + pOH = pKw = 14
pH + 2.7 = 14 and solve for pH.

c is done the same way but it's diluted; therefore (KOH) = 0.22M KOH x (10/250) = ? and go from there as in b above.

d. First determine M Ba(OH)2.
mols Ba(OH)2 = grams/molar mass, then
M = mols/L solution and proceed as in b.
Post your work if you get stuck.

Answer 2

first calcu the no. of mole of KOH

c=n/v
n=c.v =0.22*(10.0/1000) =0.0022 moles

Final conc.= 0.0022/(250/1000)= 0.0088M

pOH= -log[OH]= -log[0.0088M]=2.06
PH=14-2.06= 11.94