South Pasadena • Honors Chemistry
Name
4 · Salts and Solutions
Period
4.4
Date
QUICK CHECK
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 Strong Acid/Base, and Soluble Salt Solutions
1. Given a 0.0010 M solution of barium hydroxide. Find the pH and
pOH.
Ba(OH)2 (aq)  Ba2+ (aq) + 2 OH− (aq)
Initial
0.0010 M
0M
0M
Change
−0.0010 M
+0.0010 M
+0.0020 M
Final
0M
0.0010 M
0.0020 M
[Ba(OH)2] = 0 M
pOH = − log(0.0020) = 2.70
[Ba2+] = 0.0010 M
pH = 14.00 – 2.70 = 11.30
[OH−] = 0.0020 M
[H+] = 10−11.30 = 5.0 × 10−12 M
2. Given a 0.0010 M solution of aluminum acetate. Find the
concentration of all ions.
Al(C2H3O2)3 (s)  Al3+ (aq) + 3 C2H3O2− (aq)
Initial
0.0010 M
0M
0M
Change
−0.100 M
+0.0010 M
+0.0030 M
Final
0M
0.0010 M
0.0030 M
[Al(C2H3O2)3] = 0 M
[Al3+] = 0.0010 M
[C2H3O2−] = 0.0030 M
 Weak Acid/Base
3. Given a 0.0010 M solution of hypobromous acid. Find the pH and %
dissociation. Ka = 2 × 10−9.
HOBr (aq)  H+ (aq) + OBr− (aq)
Initial
0.0010 M
0M
0M
Change
−x
+x
+x
Equil.
0.0010 – x
x
x
[H+][OBr−]
(x)(x)
x2
Ka =
=
=
= 2 × 10−9
[HOBr]
(0.0010 − x) 0.0010
Because this is a weak acid, we can assume that x is very small
compared to 0.0010, so (0.0010 – x) ≈ 0.0010.
x = 1.4 × 10−6
[HOBr] = 0.0010 – x = 0.0010 – (1.4 × 10−6) = 0.0010 M
[H+] = x = 1.4 × 10−6 M
[OBr−] = x = 1.4 × 10−6 M
−6
pH = − log (1.4 × 10 ) = 5.85
x
1.4 × 10−6 M
% dissociation = × 100% =
× 100% = 0.14%
i
0.0010 M
 Insoluble Salt
4. Given a saturated solution of silver chromate. Find its solubility and
the concentration of dissolved ions. Ksp = 9.0 × 10−12.
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