1. Calculate pH of the following solutions: a. (H'] = 0.00001M b. (H'] = 2.30 x 10°M c. 0.05M NaOH, 100% ionized d. 0.02M HC,H3O2, a weak acid with Ka = 1.8 x 10 e. (OH'] = 3.5 x 10* M

Chemistry: Principles and Reactions
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Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 21QAP: Calculate the pH of a solution prepared by mixing 49.0 mL of butyric acid, HC4H7O2, with 6.15 g of...
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Calculate pH of the following solutions: 
a. [H+] = 0.00001M 
b. [H+] = 2.30 x 10-9
c. 0.05M NaOH, 100% ionized 
d. 0.02M HC2H3O2, a weak acid with Ka = 1.8 x 10-5
e. [OH-] = 3.5 x 10-8


The other picture are the original question and the other is example 
im sorry lack of pictures 2 is limit hehe

Calculate pH of the following solutions:
a. [H'] = 0.00001M
b. (H'] = 2.30 x 10° M
0.05M NaOH, 100% ionized
d. 0.02M HC,H;O2, a weak acid with Ka = 1.8 x 105
C.
e. [OH'] = 3.5 x 10° M
1.
Transcribed Image Text:Calculate pH of the following solutions: a. [H'] = 0.00001M b. (H'] = 2.30 x 10° M 0.05M NaOH, 100% ionized d. 0.02M HC,H;O2, a weak acid with Ka = 1.8 x 105 C. e. [OH'] = 3.5 x 10° M 1.
Example 4.1
Calculate the pH of each of the following solutions:
a. [H'] = 2.5 x 1os
b. [OH] = 2.0 x 10%
РОН 3 4.2
d. 0.02M HCI, 100% ionized
e. 0.10M HCN, weak acid with Ka= 4.0 x 1010
C.
Solution:
pH = - log(H'] = - log (2.5 x 10) = - (-4.6) = 4.6
a.
pOH = - log (OH'] = - log (2.0 x 10®) = -(-7.69) = 7.69
pH + pOH = 14
рH 3D14 - рОН
b.
pH = 14 – 7.69 = 6.31
РОН 3 4.2
C.
pH + pОH %3D 14
pH 3D 14 — рОН
= 14 – 4.2
= 9.8
d.
0.02M HCI 100% ionized will have [H*] = 0.02M
pH = - log (H'] = - log (0.02) = - (-1.69) = 1.69
Transcribed Image Text:Example 4.1 Calculate the pH of each of the following solutions: a. [H'] = 2.5 x 1os b. [OH] = 2.0 x 10% РОН 3 4.2 d. 0.02M HCI, 100% ionized e. 0.10M HCN, weak acid with Ka= 4.0 x 1010 C. Solution: pH = - log(H'] = - log (2.5 x 10) = - (-4.6) = 4.6 a. pOH = - log (OH'] = - log (2.0 x 10®) = -(-7.69) = 7.69 pH + pOH = 14 рH 3D14 - рОН b. pH = 14 – 7.69 = 6.31 РОН 3 4.2 C. pH + pОH %3D 14 pH 3D 14 — рОН = 14 – 4.2 = 9.8 d. 0.02M HCI 100% ionized will have [H*] = 0.02M pH = - log (H'] = - log (0.02) = - (-1.69) = 1.69
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