What is the pH of a 0.02M solution of potassium benzoate solution ( K+ -OOCC6H5). The pka of benzoic acid is 4.19. (Write answer to the hundredths place)
use:
pKa = -log Ka
4.19 = -log Ka
Ka = 6.457*10^-5
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/6.457*10^-5
Kb = 1.549*10^-10
C6H5COO- dissociates as
C6H5COO- + H2O -----> C6H5COOH + OH-
0.02 0 0
0.02-x x x
Kb = [C6H5COOH][OH-]/[C6H5COO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.549*10^-10)*2*10^-2) = 1.76*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.76*10^-6 M
use:
pOH = -log [OH-]
= -log (1.76*10^-6)
= 5.7545
use:
PH = 14 - pOH
= 14 - 5.7545
= 8.2455
Answer: 8.25
What is the pH of a 0.02M solution of potassium benzoate solution ( K+ -OOCC6H5). The...
What is the pH of a 0.02M solution of potassium benzoate solution ( K+ -OOCC6H5). The pka of benzoic acid is 4.19. (Write answer to the hundredths place) please help me
Question 3 What is the pH of a 0.02M solution of potassium benzoate solution (K-OOCC6H5). The pka of benzoic acid is 4.19. (Write answer to the hundredths place)
What is the pH of a 0.02 M solution of potassium benzoate solution (K+ -OOCC6H5). The pka of benzoic acid is 4.19. (Write answer to the hundredths place). What is the pH of a 0.02M solution of potassium benzoate solution (K+ -OOCC6H5). The pka of benzoic acid is 4.19. (Write answer to the hundredths place)
A benzoic acid/potassium benzoate buffer solution has a pH = 4.25. The concentration of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5) in the solution is 0.54 M. What is the concentration of potassium benzoate (KC6H5COO, Kb = 1.5 × 10-10) in this buffer solution?
A benzoic acid/potassium benzoate buffer solution has a pH = 4.25. The concentration of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5) in the solution is 0.54 M. What is the concentration of potassium benzoate (KC6H5COO, Kb = 1.5 × 10-10) in this buffer solution?
, and 0.41 M in potassium benzoate A solution is prepared at 25 °C that is initially 0.14 M in benzoic acid (HC H,CO2), a weak acid with K -6.3 10 (KCH,CO2). Calculate the pH of the solution. Round your answer to 2 decimal places. pH-
The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium benzoate and 50.0 mL of 1.00 M benzoic acid is The pka of benzoic acid is 4.20. 1.71 3.41 2.38 4.20 0.85
Calculate the pH at 25°C of a 0.54M solution of sodium benzoate NaC6H5CO2. Note that benzoic acid HC6H5CO2 is a weak acid with a pKa of 4.20.Round your answer to 1 decimal place.
What is the pH of a 0.0100 M solution of sodium benzoate, NaC7H302? (K for benzoic acid, HC7H502 - 6.3 x 10-5). CyH5O2 (aq) + H20(1) = HC7H5O2(aq) + OH(aq) O pH = 9.82 O pH - 0.38 O pH = 5.91 O pH = 8.10 O pH = 13.62
What is the pH of a 0.1M solution of benzoic acid? ( Write answer to the hundredths place) KA(benzoic acid) = 6.46x10-5