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
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.