6. You make a carbonate buffer by adding 4.00g sodium carbonate and 3.50g sodium
ID: 1074271 • Letter: 6
Question
6. You make a carbonate buffer by adding 4.00g sodium carbonate and 3.50g sodium bicarbonate to 300mL water and then bring the solution up to 650mL. (Please show your work for me!) A) carbonic acid has 2 Kas: 4.30 x 10^-7 and 5.61 x10^-11. Which is the relevant Ka, why? B) what is the pH of this solution? C) what is the pH of the solution after adding 10 mmol HCl? Dase in the bufeer to form weat ac littl e char M sodium carbonate and 3.50 g sodium bicarbonate to 6. You make a carbonate buffer by adding 4.00 9 a fe CH result 1 n 9 You make a carbonate acid has two K's: 4.30 x 10i and 5.61 x 10-11, Which is the relevant K.? Why? 0.13 M Na aco3 a. Carbonic acid then bring the solution up to 650 0 mL b. What is the pH of this solution? What is the pH of this solution after adding 10 mmol HCI? c. d. What is the pH of this solution after adding 10 mmol NaOH?Explanation / Answer
a) Ka2 is the relevant Ka
This buffer is mixture of acid HCO3- and conjugate base CO32-
HCO3- <-------> CO32- + H+
Ka2 = [CO32-][H+] /[HCO3-] = 5.61*10-11
b) no of mole of sodium bicarbonate = 3.50g/84.0066g/mol=0.04163mol
[HCO3-] = (0.04163mol/650ml)*1000ml = 0.0640M
No of mole of CO32- = 4g/105.989g/mol =0.03774M
[CO32-] = (0.03774mol/650ml)*1000ml=0.0581M
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
= 10.25 + log(0.0581/0.0640)
= 10.25 - 0.04
= 10.21
c) 10mmol= 0.010mol
HCl react with the conjugate base CO32-
HCl + CO32- --------> HCO3- + Cl-
0.010mole of HCl react with 0.010mol of CO32- to give 0.010mole of HCO3-
After adding
No of mole of CO32- = 0.03774 - 0.010 = 0.02774mole
No of mole of HCO3- = 0.04163 + 0.010 = 0.05163
[HCO3-] = (0.05163mol/650ml)*1000ml=0.0794M
[CO32-] = ( 0.02774mol/650ml)*1000ml =0.0427M
Applying Henderson-Hasselbalch equation
pH = 10.25 + log(0.0427/0.0794)
= 10.25 - 0.27
= 9.98
d) NaOH react with Acid HCO3-
HCO3- + OH- - - - - - > CO32- + H2O
After adding
No of mole of HCO3- = 0.04163 - 0.010 =0.03163M
No of mole of CO32- = 0.03774 + 0.010 = 0.04774M
[CO32-] = 0.0734M
[HCO3-] = 0.0487M
Applying Henderson - Hasselbalch equation
pH = 10.25 + log(0.0734/0.0487)
= 10.25 + 0.18
= 10.43
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