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Calculation of pH for Various Concentrations of Ba(OH)2, HNO2, and Acetic Acid Solutions, Lecture notes of Chemistry

Step-by-step calculations for determining the ph of solutions with different concentrations of ba(oh)2, hno2, and acetic acid using various methods. It covers strong and weak bases and acids, and the use of the henderson-hasselbalch equation for buffer solutions.

What you will learn

  • How to calculate the pH of a weak acid like HNO2?
  • What is the pH of a solution that is 0.080 M in acetic acid and 0.160 M in sodium acetate?
  • What is the pH of a 0.020 M Ba(OH)2 solution?

Typology: Lecture notes

2021/2022

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CHEM1002 2008-N-5 November 2008
โ€ข Calculate the pH of a 0.020 M solution of Ba(OH)2.
Marks
1
Ba(OH)2 is a strong base so it will completely dissociate in solution:
Ba(OH)2(s) ร ๏ƒ  Ba2+(aq) + 2OH-(aq)
As each Ba(OH)2 dissociates to make 2OH-, a 0.020 M solution has [OH-(aq)] =
0.040 M.
By definition, pOH = -log10[OH-(aq)] so pOH = -log10(0.040) = 1.40.
As pH + pOH = 14.00,
pH = 14.00 โ€“ 1.40 = 12.60
pH = 12.60
โ€ข Calculate the pH of a 0.150 M solution of HNO2. The pKa of HNO2 is 3.15.
3
As HNO2 is a weak acid, [H3O+] must be calculated using a reaction
table:
HNO2
H2O
H3O+
NO2-
initial
0.150
large
0
0
change
-x
negligible
+x
+x
final
0.150 โ€“x
large
x
x
The equilibrium constant Ka is given by:
Ka =
๐‡๐Ÿ‘๐Ž![๐๐Ž๐Ÿ
!]
[๐‡๐๐Ž๐Ÿ] =
๐’™๐Ÿ
๐ŸŽ.๐Ÿ๐Ÿ“๐ŸŽ!๐’™
As pKa = -log10Ka, Ka = 10โˆ’3.15 and is very small, 0.150 โ€“ x ~ 0.150 and hence:
x2 = 0.150 ร— 10โ€“3.15 or x = 1.03 ร— 10-2 M = [H3O+]
Hence, the pH is given by:
pH = โˆ’log10[H3O+] = โˆ’log10(1.03 ร— 10-2) = 1.987
pH = 1.987
ANSWER CONTINUES ON THE NEXT PAGE
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CHEM1002 2008 - N- 5 November 2008

  • Calculate the pH of a 0.020 M solution of Ba(OH) 2. Marks 1 Ba(OH) 2 is a strong base so it will completely dissociate in solution: Ba(OH) 2 (s) ร ๏ƒ  Ba2+(aq) + 2OH-(aq) As each Ba(OH) 2 dissociates to make 2OH

, a 0.020 M solution has [OH

(aq)] = 0.040 M. By definition, pOH = - log 10 [OH-(aq)] so pOH = - log 10 (0.040) = 1.40. As pH + pOH = 14.00, pH = 14.00 โ€“ 1.40 = 12. pH = 12.

  • Calculate the pH of a 0.150 M solution of HNO 2. The p K a of HNO 2 is 3.15. 3 As HNO 2 is a weak acid, [H 3 O + ] must be calculated using a reaction table: HNO 2 H 2 O H 3 O

NO 2

initial 0.150 large 0 0 change - x negligible + x + x final 0.150 โ€“ x large x x The equilibrium constant K a is given by: K a = ๐‡๐Ÿ‘๐Ž!^ [๐๐Ž๐Ÿ!] [๐‡๐๐Ž๐Ÿ]

๐’™๐Ÿ ๐ŸŽ.๐Ÿ๐Ÿ“๐ŸŽ!๐’™ As p K a = - log 10 K a, K a = 10โˆ’^ 3.15^ and is very small, 0.150 โ€“ x ~ 0.150 and hence: x 2 = 0.150 ร— 10

    or x = 1.03 ร— 10 - 2 M = [H 3 O + ] Hence, the pH is given by: pH = โˆ’log 10 [H 3 O+] = โˆ’log 10 (1.03 ร— 10 -^2 ) = 1. pH = 1.9 87 ANSWER CONTINUES ON THE NEXT PAGE

CHEM1002 2008 - N- 5 November 2008

  • Calculate the pH of a solution that is 0.080 M in acetic acid and 0.160 M in sodium acetate. The pKa of acetic acid is 4.76.

The solution contains both a weak acid (acetic acid) and its conjugate base (the acetate ion). This is a buffer and its pH can be calculate using the Henderson- Hasselbalch equation. With [acid] = 0.080 M and [base] = 0.160 M, pH = p K a + log [๐›๐š๐ฌ๐ž] [๐š๐œ๐ข๐] = 4.76 + log ๐ŸŽ.๐Ÿ๐Ÿ”๐ŸŽ ๐ŸŽ.๐ŸŽ๐Ÿ–๐ŸŽ

pH = 5.