A 10.0-mL sample of 1.0 M NaHCO 3 is titrated with 1.0 M HCl 1. What is the pH a
ID: 943436 • Letter: A
Question
A 10.0-mL sample of 1.0 M NaHCO3 is titrated with 1.0 M HCl
1. What is the pH after 0 mL HCl added?
2. What is the pH after 1.0 mL HCl added?
3. What is the pH after 9.5 mL HCl added?
4. What is the pH after 10.0 mL HCl added (equivalence point)?
5. What is the pH after 10.5 mL HCl added?
6. What is th pH after 12.0 mL HCl added?
PLEASE SHOW THE WORK SO I CAN FIGURE OUT HOW TO DO THESE...
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When creating a titration curve for a weak base, the pH of the initial solution requires setting up a table showing the initial, change, and equilibrium values for each species and plugging these into the equilibrium constant expression. The expression for the bicarbonate ion is H2COs For an anion that can both hydrolyze and produce HF, the pH of a concentrated solution can be more easily approximated using the equation During the titration before the equivalence point, provided that the concentration of acid is significantly more than the concentration of base, the Henderson-Hasselbalch equation can be used to approximate the pH: base (acid] At the equivalence point, the solution is no longer a buffer, but contains the weak acid H2CO3. The change in concentration of the bicarbonate ion is significant, and the equilibrium constant expression for Kal must again be used to find the concentration of hydronium ions. After the equivalence point, the strong acid will control the pH.Explanation / Answer
1. What is the pH after 0 mL HCl added?
HCO3 - + H2O -----------------> H2CO3 + OH-
1.0 0 0 --------------> initial
1.0-x x x --------------> equilibrium
Kb1 = Kw / Ka1 = [H2CO3][OH-]/[HCO3-]
(1.0x 10^-14) / (4.43 x 10^-7) = x^2 / 1-x
2.26 x 10^-8 = x^2 / 1-x
x^2 + 2.26 x 10^-8 x - 2.26 x 10^-8 = 0
by solving this : x = 1.50 x 10^-4
[OH-] = 1.50 x 10^-4 M
pOH = -log [OH-] = 3.82
pH + pOH = 14
pH = 10.18
2. What is the pH after 1.0 mL HCl added?
millimoles of HCO3 - = 10 x 1 = 10
millimoles of HCl = 1 x 1 = 1
HCO3- + H+ ------------------> H2CO3
10 1 0
9 0 1
pH = pKa1 + log [HCO3-] / [H2CO3]
pH = 6.35 + log (9/1)
pH = 7.30
3. What is the pH after 9.5 mL HCl added?
millimoles of HCO3 - = 10 x 1 = 10
millimoles of HCl = 1 x 9.5 = 9.5
HCO3- + H+ ------------------> H2CO3
10 9.5 0
0.5 0 9.5
pH = pKa1 + log [HCO3-] / [H2CO3]
pH = 6.35 + log (0.5 /9.5)
pH = 5.07
4. What is the pH after 10.0 mL HCl added (equivalence point)?
millimoles of HCO3 - = 10 x 1 = 10
millimoles of HCl = 1 x 10 = 9.5
HCO3- + H+ ------------------> H2CO3
10 10 0
0 0 10
only H2CO3 remains
H2CO3 concentration = 10 / (20) = 0.5 M
pH = 1/2 [pKa1 -logC]
pH = 1/2 [6.35 -log 0.5]
pH = 6.35 + log (0.5 /9.5)
pH = 3.02
5. What is the pH after 10.5 mL HCl added?
HCO3- + H+ ------------------> H2CO3
10 10.5 0
0 0.5 10
here strong acid remains . so pH decided by stong acid
[H+] concentration = 0.4 / (10 + 10.5) = 0.0195 M
pH = -log [H+]
pH = 1.71
6. What is th pH after 12.0 mL HCl added?
HCO3- + H+ ------------------> H2CO3
10 12 0
0 2 10
here strong acid remains . so pH decided by stong acid
[H+] concentration = 2 / (10 + 12) = 0.09 M
pH = -log [H+]
pH = 1.04
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