Question

In: Chemistry

Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5)...

Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH).

A. 8.56

B. 7.95

C. 5.33

D. 6.25

Solutions

Expert Solution

we have:

Molarity of C6H5COOH = 0.2 M

Volume of C6H5COOH = 15 mL

Molarity of NaOH = 0.15 M

Volume of NaOH = 20 mL

mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH

mol of C6H5COOH = 0.2 M * 15 mL = 3 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.15 M * 20 mL = 3 mmol

We have:

mol of C6H5COOH = 3 mmol

mol of NaOH = 3 mmol

3 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base

C6H5COO- formed = 3 mmol

Volume of Solution = 15 + 20 = 35 mL

Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.5*10^-5 = 1.538*10^-10

concentration ofC6H5COO-,c = 3 mmol/35 mL = 0.0857M

C6H5COO- dissociates as

C6H5COO- + H2O -----> C6H5COOH + OH-

0.0857 0 0

0.0857-x x x

Kb = [C6H5COOH][OH-]/[C6H5COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.538*10^-10)*8.571*10^-2) = 3.631*10^-6

since c is much greater than x, our assumption is correct

so, x = 3.631*10^-6 M

[OH-] = x = 3.631*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.631*10^-6)

= 5.44

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.44

= 8.56

Answer: A


Related Solutions

Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5)...
Calculate the pH when 15.0 mL of 0.20 M benzoic acid (HC7H5O2, Ka = 6.5 ×10–5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH).
Calculate the pH when 15.0 mL of 0.20 M Benzoic Acid (HC7H5O2 Ka = 6.5x10-5) is...
Calculate the pH when 15.0 mL of 0.20 M Benzoic Acid (HC7H5O2 Ka = 6.5x10-5) is titrated with 20.0 mL of 0.15 M sodium hydroxide (NaOH) Can you exaplain how to solve this question?
Calculate the values of pH when 40.0 mL of 0.0250M benzoic acid (HC7H5O2 , Ka=6.3x10-5 )...
Calculate the values of pH when 40.0 mL of 0.0250M benzoic acid (HC7H5O2 , Ka=6.3x10-5 ) is titrated with (1) 0.0 mL of 0.050 M NaOH solution? (2) 10.0 mL of 0.050 M NaOH solution? (3) 25.0 mL of 0.050 M NaOH solution? I mostly need help with 2 and 3. The answers are 4.20 and 11.58 but I not know how to calculate them.
The Ka of benzoic acid is 6.5 x 10 ^(-5) a) Calculate the pH of a...
The Ka of benzoic acid is 6.5 x 10 ^(-5) a) Calculate the pH of a 40.00 mL, 0.1 M benzoic acid buffer solution after the addition of 20.00 mL of a 0.1M NaOH b) Calculate the pH of a 40.00 mL, 0.1 M benzoic acid buffer solution after the addition of 50.00 mL of a 0.1M NaOH REDOX TITRATION 2) Express the reactions for potassium permanganate titration with sodium oxalate: Hint: include the 2 half-reactions. 3) What is the...
Ka for Benzoic acid (a monoprotic acid) is Ka = 6.5*10-5. Calculate the pH after addition...
Ka for Benzoic acid (a monoprotic acid) is Ka = 6.5*10-5. Calculate the pH after addition of 23.0, 35.0, 46.0, and 49.0 mL of 0.100M NaOH to 46.0 mL of 0.100M Benzoic acid
If 10 mL of 0.025 M benzoic acid (Ka = 6.5 * 10-5) is titrated with...
If 10 mL of 0.025 M benzoic acid (Ka = 6.5 * 10-5) is titrated with 0.01 M NaOH, what is pH after 0 ml, 20 ml, 50 ml, and 90 ml of base have been added?
Calculate the pH in 0.050 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5×10−5.
Calculate the pH in 0.050 M sodium benzoate; Ka for benzoic acid (C6H5CO2H) is 6.5×10−5.
Determine the pH of the solution of 100 mL of 0.150 M benzoic acid, HC7H5O2 (Ka=6.3x10-5...
Determine the pH of the solution of 100 mL of 0.150 M benzoic acid, HC7H5O2 (Ka=6.3x10-5 ). To this 25.0 mL of 0.400 M KOH is added, calculate the pH of the resulting solution. In a second step an additional 12.5ml of 0.400M KOH is added, calculate the pH of the resulting solution.
Determine the pH of a 0.150 M benzoic acid solution with a  Ka = 6.5 x 10–5.  ...
Determine the pH of a 0.150 M benzoic acid solution with a  Ka = 6.5 x 10–5.   - report the answer in three significant figures
25.0 mL of a 0.500 M solution of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5)...
25.0 mL of a 0.500 M solution of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5) is treated with 25.0 mL of a 0.400 M solution of KOH. Determine the pH of the solution. A. 4.79    B. 4.29    C. 4.09       D. 3.59
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT