Answer:
(a) pH = 12.73
(b) pH = 10.52
(c) pH = 1.93
Explanation:
The net balanced reaction equation is:
KOH + HBr ⇒ H₂O + KBr
The amount of KOH present is:
n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol
(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol
This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:
(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH
After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:
C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH
The pOH and pH of the solution can then be calculated:
pOH = -log[OH⁻] = -log(0.0538461) = 1.2688
pH = 14 - pOH = 14 - 1.2688 = 12.73
(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol
This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:
(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH
After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:
C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH
The pOH and pH of the solution can then be calculated:
pOH = -log[OH⁻] = -log(0.0003344) = 3.476
pH = 14 - pOH = 14 - 3.476 = 10.52
(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:
(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol
This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:
(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr
After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:
C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr
The pH of the solution can then be calculated:
pH = -log[H⁺] = -log(0.01176) = 1.93
......(1)
.......(2)