(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.13E
The
Explanation of Solution
The
If consider,
Therefore, the
(b)
Interpretation:
The
Concept introduction:
Refer to part (a).
(b)
Answer to Problem 6D.13E
The
Explanation of Solution
The equilibrium reaction of methylamine 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.
Methylamine
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,
Now, the
Therefore, the calculated
(c)
Interpretation:
The
Concept introduction:
Refer to part (a).
(c)
Answer to Problem 6D.13E
The
Explanation of Solution
Acetic acid is a weak acid 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.
Acetic acid
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,
Now, the
Therefore, the calculated
(d)
Interpretation:
The
Concept introduction:
Refer to part (a).
(d)
Answer to Problem 6D.13E
The
Explanation of Solution
Aniline 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 are obtained in the above table and is substituted in above equation and is given below.
Aniline
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 concentration of
The
Therefore, the calculated
Using the
Therefore, the
The solutions are ranked in the order of increasing pH and it is given below.
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Chapter 6 Solutions
Chemical Principles: The Quest for Insight
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- Calculate the pH of 0.10 M (COOH)2 (aq), oxalic acid. Ka1 = 5.9 × 10–2 ; Ka2 = 6.4 × 10–5.arrow_forwardEstimate the pH of 0.05 M NaCN(aq) at 25°C. Ka(HCN) = 3.4 × 10−9. Write your answer with two decimal placesarrow_forwardCalculate the pH of a solution by dissolving 1.90 g of sodium acetate, CH3COONa, in 71.0 mL of 0.15 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. Last of CH3COOH is 1.75 x 10^-5arrow_forward
- Calculate pH of 2.5 x 10-2 M NaOH(aq). NaOH(aq) is strong base.arrow_forwardDetermine the pH of a 0.450M solution of aniline hydrochloride. C6H;NH, (ag) + H,O(1) = H;O* (ag) + C,H;NH2(aq)arrow_forwardFor the following acid-base reaction, (1) predict the products, showing both reactants and products complete Lewis structures and arrows showing electron flow; (2) label each structure with the lowing: Bronsted acid, Bronsted base, conjugate acid, conjugate base; (3) give a brief definition of a ronsted acid and Bronsted base; (4) predict the direction of the equilibrium and justify your answer. HC0OH + CH3 Nta PRん106Y pkb = 3.36arrow_forward
- Predict whether aqueous solutions of the following substances are acidic, basic, or neutral and write hydrolysis equations for the acidic and basic solutions. (a) CsBr; (b) Al(NO3)3; (c) KCN; (d) CH3NH3Clarrow_forwardCalculate the pH of the resulting solution if 22.0 mL of 0.220 M HCl(aq) is added to 32.0 mL of 0.220 M NaOH(aq). pH = Calculate the pH of the resulting solution if 22.0 mL of 0.220 M HC1(aq) is added to 12.0 mL of 0.320M NAOH(aq). pH =arrow_forwardCalculate the pH of the resulting solution if 26.0 mL of 0.260 M HCl(aq) is added to 36.0 mL of 0.260 M NaOH(aq). pH = Calculate the pH of the resulting solution if 26.0 mL of 0.260 M HCI(aq) is added to 16.0 mL of 0.360 M NaOH(aq). pH IIarrow_forward
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