Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Calculate the pH of 0.310L of a 0.495M NaHSO3 - 0.315MNa2SO3 buffer before and a

ID: 838911 • Letter: C

Question

Calculate the pH of 0.310L of a 0.495M NaHSO3 - 0.315MNa2SO3 buffer before and after the addition of the following substances.

Part A

Calculate the initial pH.

Express your answer using two decimal places.

SubmitMy AnswersGive Up

Part B

Calculate the pH after the addition of 5.0 mL of 0.170M HCl

Express your answer using two decimal places.

SubmitMy AnswersGive Up

Part C

Calculate the pH after the addition of 5.0 mL of 0.140M NaOH.

Express your answer using two decimal places.

Calculate the pH of 0.310L of a 0.495M NaHSO3 - 0.315MNa2SO3 buffer before and after the addition of the following substances.

Part A

Calculate the initial pH.

Express your answer using two decimal places.

pHinitial =

SubmitMy AnswersGive Up

Part B

Calculate the pH after the addition of 5.0 mL of 0.170M HCl

Express your answer using two decimal places.

pH =

SubmitMy AnswersGive Up

Part C

Calculate the pH after the addition of 5.0 mL of 0.140M NaOH.

Express your answer using two decimal places.

pH =

Explanation / Answer

pH = pKa (of NaHSO3) + log ([Na2SO3]/[NaHSO3])

NOTE: you can use the molarity of Na2SO3 and NaHSO3 or the moles of Na2SO3 and NaHSO3 since this is a ratio and moles/L over moles/L = moles over moles.
The Ka for NaHSO3 = 6.30x10^-8 so the pKa = -log(6.30x10^-8) = 7.200

Before addition of HCl or NaOH
pH = 7.200 + log (0.315 / 0.495) = 7.200 - 0.1963 = 7.0037 .....rounding to two decimals 7.00

0.310 L x 0.485 M = 0.15035 mole NaHSO3
0.310 L x 0.315 M = 0.09765 mole Na2SO3
0.005 L x 0.14 M HCl = 0.0007 mole HCl
0.005 L x 0.14 M NaOH = 0.0007 mole NaOH

After addition of HCl

moles of NaHSO3 = 0.15035 mole + 0.0007 mole = 0.15105 mole
moles of Na2SO3 = 0.09765 mole - 0.0007 mole = 0.09695 mole
pH = 7.20 + log (0.09695 / 0.15105) = 7.200 - 0.1926 = 7.007 ..... rounding to two decimals 7.01

After addition of NaOH
moles of NaHSO3 = 0.15035 mole - 0.0007 mole = 0.14965 mole
moles of Na2SO3 = 0.09765 mole + 0.0007 mole = 0.09835 mole
pH = 7.20 + log (0.09835 / 0.14965) = 7.200 - 0.1823 = 7.0177 .....rounding to two decimals 7.02

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote