Solve the following problems:
1. pH of an aqueous solution of potassium hydroxide at 298 K temperature is 11.65. The initial volume of this solution is made 6 times by addition of water. What will be the pH of the diluted solution ?
2. What will be the change in the value of pH if the concentration of the aqueous solution of HNO3is increased to 0.05 M from 0.03 M ?
1. pH = 11.65
Therefore, pOH = 14 – 11.65 = 2.35
Using, pOH = -log10[OH-]
Putting the value of pOH in above equation:
2.35 = -log10[OH-]
log10[OH-] = -2.355
log10[OH-] = 3 -2.35 -3
log10[OH-] = -3 + 0.65
log10[OH-] =
+ 0.65
[OH-] = antilog![]()
[OH-] = 4.46 × 10-3
Now if volume is made 6 times then the concentration will decrease 6 times. So the new concentration is = 4.46× 10-3÷ 6
= 7.44 × 10-4
So new pOH = -log10[OH-]
= -log10[7.44 × 10-4]
= 4-log10[7.44]
= 4- 0.871
= 3.129
Therefore, new pH is 14 – 3.129 = 10.871
2. A) when concentration is 0.03 M
Using, pH = -log10[H3O + ]
= -log10[0.03]
= -log10[3 × 10-2]
= 2 -log10[3]
= 2 – 0.477
= 1.52
B) When concentration is 0.05 M
Using, pH = -log10[H3O + ]
= -log10[0.05]
= -log10[5 × 10-2]
= 2 -log10[5]
= 2 – 0.699
= 1.30
So the change in pH is = 1.52 – 1.30
= 0.22
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