Concept explainers
Calculate the pH of a 0.20 M NaHCO3 solution.(Hint: As an approximation, calculate hydrolysis and ionization separately first, followed by partial neutralization.)

Interpretation:
The 0.20 MNaHCO3 solution pH range has to be calculated
Concept Information:
Acid ionization constant Ka:
The equilibrium expression for the reaction HA(aq)⇌H+(aq)+A-(aq) is given below.
Ka = [H+][A-][HA]
Where Ka is acid ionization constant, [H+] is concentration of hydrogen ion, [A-] is concentration of acid anion, [HA] is concentration of the acid
Base ionization constant Kb
The equilibrium expression for the ionization of weak base B will be,
B(aq)+H2O(l) ⇌ HB+(aq)+OH-(aq)
Kb=[HB+][OH-][B]
Where Kb is base ionization constant, [OH−] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base
Relationship between Kaand Kb
Ka×Kb=Kw
pH definition:
The concentration of hydrogen ion is measured using pH scale. The acidity of aqueous solution is expressed by pH scale.
The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.
pH = -log[H3O+]
Answer to Problem 15.157QP
The sodium bicarbonate NaHCO3 pH range is 9.82
Explanation of Solution
First we write the two separate equilibrium reactions with hydrogen carbonate functioning as basic as an acid in the other.
The hydrogen ions and hydroxide ions produced in the equilibria will then undergo partial neutralization.
The Kb chemical equilibrium denoted by
HCO−3(aq)+ H2O(l)⇌H2CO3(aq)+ OH-(aq) | |||
Initial (M) |
0.20 -x 0.20−x | 0.00 | 0.00 |
Change (M) | +x | +x | |
Equilibrium (M) | x | x |
The kb value HCO3= 2.4×10−8
The equilibrium expression is,
Kb = [H2CO3][OH−][HCO−3]2.4×10−8 = x2(0.20−x)
We assume that x is small so, (0.20−x) ≈ 0.20
2.4 ×10−8 = x2(0.20)x= [OH] =6.9×10−5 M
The Ka chemical equilibrium denoted by
HCO−3(aq)⇌H+(aq)+ CO3-(aq) | |||
Initial (M) |
0.20 -x 0.20−x | 0.00 | 0.00 |
Change (M) | +x | +x | |
Equilibrium (M) | x | x |
The ka value HCO3= 4.8×10−11
The equilibrium expression is,
Ka = [H+][CO2−3][HCO−3]4.8×10−11 = x2(0.20−x)
We assume that x is small so, (0.20−x) ≈ 0.20
4.8×10−11 = x2(0.20)x= [H+] =3.1×10−6 M
The hydroxide OH− produced in the first equilibrium will be partially neutralized by the H+ produced in the second equilibrium.
H+(aq) + OH-(aq) ⇌H2O(l) | |||
Initial (M) |
3.1×10−6 −3.1×10−6 0 | 6.9×10−5 | 0.00 |
Change (M) | 3.1×10−6 | 3.1×10−6 | |
Equilibrium (M) | +6.6×10−5 | 3.1×10−6 |
The Kb value of hydroxide is a 6.6×10−5 M
Calculation of pOH value
pOH = log (6.6×10-5)pOH = 4.18_Calculation of pH valuepH= pOH− 4.18pH= 14-4.18pH= 9.82_
Therefore given NaHCO3 solution pOH = 4.18 and pH= 9.82
The pOH and pH result is only an approximation to the more rigorous treatment.
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Chapter 15 Solutions
AVC LOOSELEAF CHEMISTRY W/CONNECT 2 SEM
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