(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.
Want to see more full solutions like this?
Chapter 6 Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
- Don't used hand raiting and don't used Ai solutionarrow_forwardDraw the condensed structure of 4-ethyl-1,2,4-trifluoro-2-methyloctane.arrow_forward5. The existence of compounds of the noble gases was once a great surprise and stimulated a great deal of theoretical work. Label the molecular orbital diagram for XeF (include atom chemical symbol, atomic orbitals, and molecular orbitals) and deduce its ground state electron configuration. Is XeF likely to have a shorter bond length than XeF+? XeF XeF+ Bond Orderarrow_forward
- Don't used hand raitingarrow_forward4. The superoxide ion, Oz, plays an important role in the ageing processes that take place in organisms. Judge whether O2 is likely to have larger or smaller dissociation energy than O2. Molecular Orbital Diagram 8 02 02 Does O2 have larger or smaller dissociation energy?: Bond Orderarrow_forwardWill a weak base with a pKa of 8.4 be best absorbed in the stomach or the intestine? Explain your reasoning behind your answer. Use a chemical equilibrium equation in your answer.arrow_forward
- You have started a patient on a new drug. Each dose introduces 40 pg/mL of drug after redistribution and prior to elimination. This drug is administered at 24 h intervals and has a half life of 24 h. What will the concentration of drug be after each of the first six doses? Show your work a. What is the concentration after the first dose? in pg/mL b. What is the concentration after the second dose? in pg/mL c. What is the concentration after the third dose? in pg/mLarrow_forwardHow many different molecules are drawn below?arrow_forwardOnly 100% sure experts solve it correct complete solutions need to get full marks it's my quiz okkkk.take your time but solve full accurate okkk chemistry expert solve it.qno4arrow_forward
- Nonearrow_forwardA complete tensile test was performed on a magnesium specimen of 12 mm diameter and 30 mm length, until breaking. The specimen is assumed to maintain a constant volume. Calculate the approximate value of the actual stress at breaking. TABLE. The tensile force F and the length of the specimen are represented for each L until breaking. F/N L/mm 0 30,0000 30,0296 5000 10000 30,0592 15000 30,0888 20000 30,15 25000 30,51 26500 30,90 27000 31,50 26500 32,10 25000 32,79arrow_forwardNonearrow_forward
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning