Calculate the equilibrium concentration of carbonate in a 0.067 M solution of ca

Question

Calculate the equilibrium concentration of carbonate in a 0.067 M solution of carbonic acid. Do not enter units as part of your answer. Use “E” for scientific notation.

Ascorbic acid (Vitamin C, CHO) is a weak diprotic acid, with K = 6.8x10 and K = 2.7x10. What is the pH of a solution that contains 10.0 mg of vitamin C per 1.0 mL of solution?
Calculate the equilibrium concentration of carbonate in a 0.067 M solution of carbonic acid. Do not enter units as part of your answer. Use “E” for scientific notation.

Ascorbic acid (Vitamin C, CHO) is a weak diprotic acid, with K = 6.8x10 and K = 2.7x10. What is the pH of a solution that contains 10.0 mg of vitamin C per 1.0 mL of solution?


Ascorbic acid (Vitamin C, CHO) is a weak diprotic acid, with K = 6.8x10 and K = 2.7x10. What is the pH of a solution that contains 10.0 mg of vitamin C per 1.0 mL of solution?

Explanation / Answer

Solution:

Carbonic acid:

The CO2 in air is normally 300 ppm, so dissolved concentration of CO2 in water will be 0.067*3*10-4=10-5 E, so we get equilibrium concentration as below,

Keq = [H+][HCO3-]/[H2CO3] = 4.3*10-7

Ascorbic acid:

molar mass of ascorbic acid = 176.12 g/mol

mass = 10 mg = 0.01 g

moles = 0.01/176.12 = 0.000057 moles

c = 0.000057/0.001 = 0.057 M

[H+] = (K*c)0.5

for K=6.8*10

[H+] = (6.8*10*0.057)0.5=1.968

pH=-log[H+] =-log(1.968)=0.294

Note: Kindly check K value which is given in question the power value of 10 is missing in the data.

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