(a)
Interpretation:
The
pH definition:
The concentration of hydrogen ion is measured using
The
On rearranging, the concentration of hydrogen ion
(a)
Answer to Problem 6D.14E
The
Explanation of Solution
The
If consider,
Therefore, the
Therefore, the
(b)
Interpretation:
The
Concept introduction:
Refer to part (a).
(b)
Answer to Problem 6D.14E
The
Explanation of Solution
The equilibrium reaction of sodium nitrite is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.20 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.20-x | x | x |
The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.
Nitrous acid
Therefore, the
The obtained
The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,
Therefore, the pOH concentration of sodium nitrite is
Now, the
Therefore, the
Therefore, the
(c)
Interpretation:
The
Concept introduction:
Refer to part (a).
(c)
Answer to Problem 6D.14E
The
Explanation of Solution
Ammonia is a weak base when it is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.20 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.20-x | x | x |
The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.
Ammonia
The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,
Therefore, the pOH concentration of ammonia is
Now, the
Therefore, the
Therefore,
(d)
Interpretation:
The
Concept introduction:
Refer to part (a).
(d)
Answer to Problem 6D.14E
The
Explanation of Solution
Sodium cyanide is a strong base when it is dissolved in water it ionized as positive and negative ions and it is given below.
The equilibrium expression for the above reaction is given below.
Initial concentration | 0.20 | 0 | 0 |
Change in concentration | -x | +x | +x |
Equilibrium concentration | 0.20-x | x | x |
The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.
Hydrocyanic acid
Therefore, the
The obtained
The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,
Therefore, the pOH concentration of sodium cyanide is
Now, the
Therefore, the
Therefore, the
The solutions are ranked in the order of increasing pH and it is given below.
Want to see more full solutions like this?
Chapter 6 Solutions
Chemical Principles: The Quest for Insight
- The pH of 0.40 M HF(aq) is 1.93. Calculate how the pH changes when 0.356 g of sodium fluoride is added to 50.0 mL of the solution. Assume that the volume change is small and can be neglectedarrow_forwardDetermine the pH of 0.082 M ammonia, NH3 (aq) and percent ionization of base.arrow_forwardThe pH of an aqueous solution of 0.291 M ammonium perchlorate, NH4C1O4 (aq), is This solution is (Assume that K (NH3) = 1.80 × 10-5.)arrow_forward
- What volume of 0.250 M HCl(aq) will completely react with 50.0 mL of 0.115 M NaOH(aq)? Write the balanced reaction and show the stoichiometry. Write the reaction of acetate ion in water. Predict the approximate pH of a solution with acetate ion present.arrow_forwardDetermine the pH of 0.231 M Ca(OH)2 (aq)arrow_forwardCalculate the pH of a .05 M solution of HCl(aq)?arrow_forward
- (i) Define pH in words. The strong acid HClaq has a pH value of 1, use the following equation for a strong acid: HClaq à H+aq + Cl-aq and convert the following expression to deduce the hydrogen ion concentration: pH = -log10 [H+] (ii) Use the above expression to deduce the pH of HCl (aq) given the concentration of the acid to be 4.5 mol/dm3arrow_forward(i) Define pH in words. The strong acid HClaq has a pH value of 1, use the following equation for a strong acid: HClaq à H+aq + Cl-aq and convert the following expression to deduce the hydrogen ion concentration: pH = -log10 [H+] (ii) Use the above expression to deduce the pH of HCl (aq) given the concentration of the acid to be 4.5 mol/dm3 pH =arrow_forwardA solution of 1.7×10−6 M HNO3(aq) (10.0 mL) is diluted to 10.0 L with pure deionized water at 25 °C . What is the pH of the resulting solution?arrow_forward
- Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning