Determine the pH of each of the following solutions. Express your answer to two decimal places.
- 0.15M KCHO2
2 .0.24M CH3NH3I
3 .0.16M KI
The Correct Answer and Explanation is :
To determine the pH of each solution, we need to analyze the nature of the solute and its behavior in water.
1. 0.15 M KCHO₂ (Potassium Formate):
otassium formate (KCHO₂) is a salt derived from formic acid (HCHO₂), a weak acid, and potassium hydroxide (KOH), a strong base.n aqueous solution, KCHO₂ dissociates completely into potassium ions (K⁺) and formate ions (CHO₂⁻).he formate ion can hydrolyze in water, accepting a proton to form formic acid and hydroxide ions (OH⁻), making the solution basic.
The hydrolysis reaction is:
HO₂⁻ + H₂O ⇌ HCHO₂ + OH⁻
To find the pH:
- Determine the ( K_b ) of the formate ion: ( K_b = \frac{K_w}{K_a} )
iven ( K_a ) of formic acid is ( 1.8 \times 10^{-4} ):
( K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \approx 5.56 \times 10^{-11} ) - Set up the expression for ( K_b ): ( K_b = \frac{[HCHO₂][OH⁻]}{[CHO₂⁻]} )
ssume ( x ) is the concentration of OH⁻ produced:
( 5.56 \times 10^{-11} = \frac{x^2}{0.15 – x} )
ince ( K_b ) is small, ( x ) is negligible compared to 0.15 M:
( 5.56 \times 10^{-11} = \frac{x^2}{0.15} )
( x^2 = 5.56 \times 10^{-11} \times 0.15 )
( x^2 = 8.34 \times 10^{-12} )
( x = \sqrt{8.34 \times 10^{-12}} \approx 9.13 \times 10^{-6} )
hus, [OH⁻] ≈ ( 9.13 \times 10^{-6} ) M - Calculate pOH: ( \text{pOH} = -\log[OH⁻] )
( \text{pOH} = -\log(9.13 \times 10^{-6}) \approx 5.04 ) - Determine pH: ( \text{pH} = 14 – \text{pOH} )
( \text{pH} = 14 – 5.04 = 8.96 )
herefore, the pH of 0.15 M KCHO₂ is approximately 8.96.
2. 0.24 M CH₃NH₃I (Methylammonium Iodide):
ethylammonium iodide (CH₃NH₃I) is a salt formed from methylamine (CH₃NH₂), a weak base, and hydroiodic acid (HI), a strong acid.n water, it dissociates into methylammonium ions (CH₃NH₃⁺) and iodide ions (I⁻).he methylammonium ion can donate a proton to water, acting as a weak acid:
H₃NH₃⁺ + H₂O ⇌ CH₃NH₂ + H₃O⁺
To find the pH:
- Determine the ( K_a ) of the methylammonium ion: ( K_a = \frac{K_w}{K_b} )
iven ( K_b ) of methylamine is ( 4.4 \times 10^{-4} ):
( K_a = \frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}} \approx 2.27 \times 10^{-11} ) - Set up the expression for ( K_a ): ( K_a = \frac{[CH₃NH₂][H₃O⁺]}{[CH₃NH₃⁺]} )
ssume ( x ) is the concentration of H₃O⁺ produced:
( 2.27 \times 10^{-11} = \frac{x^2}{0.24 – x} )
ince ( K_a ) is small, ( x ) is negligible compared to 0.24 M:
( 2.27 \times 10^{-11} = \frac{x^2}{0.24} )
( x^2 = 2.27 \times 10^{-11} \times 0.24 )
( x^2 = 5.45 \times 10^{-12} )
( x = \sqrt{5.45 \times 10^{-12}} \approx 7.38 \times 10^{-6} )
hus, [H₃O⁺] ≈ ( 7.38 \times 10^{-6} ) M - Calculate pH: ( \text{pH} = -\log[H₃O⁺] )