(a)
Interpretation:
The pH and pOH of 0.0356 M HI(aq) has to be calculated.
Concept Introduction:
Negative logarithm of the concentration of hydronium ion is the pH of solution. This can be given in equation form as follows;
pH=−log[H3O+]
(a)
Answer to Problem 6B.6E
The pH and pOH of 0.0356 M HI(aq) is 1.44 and 12.56 respectively.
Explanation of Solution
Hydroiodic acid is a strong acid and it completely dissociates in water. This can be represented as follows;
HI(aq)→H+(aq)+I−(aq)
One mole of hydroiodic acid dissociates to give one mole of hydrogen ion and one mole of iodide ion. Therefore, 0.0356 M of hydroiodic acid gives 0.0356 M of hydronium ions. Hence, the concentration of hydronium ion is 0.0356 M.
The pH of the hydroiodic acid solution is calculated as shown below;
pH=−log[H3O+]=−log(0.0356 M)=1.44
Hence, the pH of hydroiodic acid solution is 1.44.
The pOH of the solution can be calculated using the formula of pOH as follows:
pH+pOH=14pOH=14−pH=14−1.44=12.56
Hence, the pOH of hydroiodic acid solution is 12.56.
(b)
Interpretation:
The pH and pOH of 0.0725 M HCl(aq) has to be calculated.
Concept Introduction:
Refer part (a).
(b)
Answer to Problem 6B.6E
The pH and pOH of 0.0725 M HCl(aq) is 1.14 and 12.86 respectively.
Explanation of Solution
Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;
HCl(aq)→H+(aq)+Cl−(aq)
One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, 0.0725 M of hydrochloric acid gives 0.0725 M of hydronium ions. Hence, the concentration of hydronium ion is 0.0725 M.
The pH of the hydrochloric acid solution is calculated as shown below;
pH=−log[H3O+]=−log(0.0725)=1.14
Hence, the pH of hydrochloric acid solution is 1.14.
The pOH of the solution can be calculated using the formula of pOH as follows:
pH+pOH=14pOH=14−pH=14−1.14=12.86
Hence, the pOH of hydrochloric acid solution is 12.86.
(c)
Interpretation:
The pH and pOH of 3.46×10-3 M Ba(OH)2(aq) has to be calculated.
Concept Introduction:
Refer part (a).
(c)
Answer to Problem 6B.6E
The pH and pOH of 3.46×10-3 Ba(OH)2(aq) is 11.84 and 2.16 respectively.
Explanation of Solution
Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;
Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)
One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, 3.46×10-3 M of barium hydroxide gives 6.92×10-3 M of hydroxide ions. Hence, the concentration of hydroxide ion is 6.92×10-3 M.
The pOH of the barium hydroxide solution is calculated as shown below;
pOH=−log[OH−]=−log(6.92×10-3)=2.16
Hence, the pOH of barium hydroxide solution is 2.16.
The pH of the solution can be calculated using the formula of pH as follows:
pH+pOH=14pH=14−pOH=14−2.16=11.84
Hence, the pH of barium hydroxide solution is 11.84.
(d)
Interpretation:
The pH and pOH of 10.9 mg of KOH dissolved in 10.0 mL of solution has to be calculated.
Concept Introduction:
Refer part (a).
(d)
Answer to Problem 6B.6E
The pH and pOH of KOH(aq) is 9.23 and 4.71 respectively.
Explanation of Solution
Molarity of potassium hydroxide solution is calculated as shown below;
Molarity=number of moles of soluteVolume of solution=10.9 mg×1 g1000 mg56.1056 g⋅mol−1×10.0 mL=1.94×10−5 M
Therefore, 1.94×10−5 M of potassium hydroxide gives 1.94×10−5 M of hydroxide ions. Hence, the concentration of hydroxide ion is 1.94×10−5 M.
The pOH of the potassium hydroxide solution is calculated as shown below;
pOH=−log[OH−]=−log(1.94×10−5)=4.71
Hence, the pOH of potassium hydroxide solution is 4.71.
The pH of the solution can be calculated using the formula of pH as follows:
pH+pOH=14pH=14−pOH=14−4.71=9.23
Hence, the pH of potassium hydroxide solution is 9.23.
(e)
Interpretation:
The pH and pOH of 5.00 M NaOH(aq) after dilution from 10.00 mL to 2.50 L has to be calculated.
Concept Introduction:
Refer part (a).
(e)
Answer to Problem 6B.6E
The pH and pOH of 5.00 M NaOH(aq) is 12.30 and 1.70 respectively.
Explanation of Solution
Concentration of sodium hydroxide after dilution is calculated as shown below;
CinitialVinitial=CfinalVfinal(5.00 M)(10.00 mL)=Cfinal(2500 mL)Cfinal=(5.00 M)(10.00 mL)2500.0 mL=0.02 M
Therefore, 0.02 M of sodium hydroxide gives 0.02 M of hydroxide ions. Hence, the concentration of hydroxide ion is 0.02 M.
The pOH of the sodium hydroxide solution is calculated as shown below;
pOH=−log[OH−]=−log(0.02)=1.70
Hence, the pOH of sodium hydroxide solution is 1.70.
The pH of the solution can be calculated using the formula of pH as follows:
pH+pOH=14pH=14−pOH=14−1.70=12.30
Hence, the pH of sodium hydroxide solution is 12.30.
(f)
Interpretation:
The pH and pOH of 3.5×10-4 M HClO4(aq) after dilution from 5.00 mL to 25.0 mL has to be calculated.
Concept Introduction:
Refer part (a).
(f)
Answer to Problem 6B.6E
The pH and pOH of 3.5×10-4 M HClO4(aq) is 4.15 and 9.85 respectively.
Explanation of Solution
Concentration of perchloric acid after dilution is calculated as shown below;
CinitialVinitial=CfinalVfinal(3.5×10-4 M)(5.00 mL)=Cfinal(25.0 mL)Cfinal=(3.5×10-4 M)(5.00 mL)25.0 mL=0.7×10-4 M
Therefore, 0.7×10-4 M of perchloric gives 0.7×10-4 M of hydronium ions. Hence, the concentration of hydronium ion is 0.7×10-4 M.
The pH of the perchloric acid solution is calculated as shown below;
pH=−log[H3O+]=−log(0.7×10-4)=4.15
Hence, the pH of perchloric acid solution is 4.15.
The pOH of the solution can be calculated using the formula of pH as follows:
pH+pOH=14pOH=14−pH=14−4.15=9.85
Hence, the pOH of perchloric acid solution is 9.85.
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Chapter 6 Solutions
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