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What is the pH of the buffer that results when 0.235 mol of NH3 and (2.46x10^-1)...

What is the pH of the buffer that results when 0.235 mol of NH3 and (2.46x10^-1) mol of NH4Cl are dissolved in water to a volume of 0.50 L? (Ka of NH4(+) = 5.6E-10)

Expert Solution

Ka of NH4+ = 5.6 x 10^-10

Ka x Kb = Kw = 1.0 x 10^-14

Kb of NH4+ = 1.0 x 10^-14 / 5.6 x 10^-10

= 1.78 x 10^-5

pKb = -log Kb

= -log (1.78 x 10^-5)

= 4.75

NH3 molarity (base)= 0.235 / 0.5

= 0.47 M

NH4Cl molarity (salt)= 2.46x10^-1 / 0.5

= 0.492 M

pOH = pKb + log (salt / base)

= 4.75 + log [0.492 / 0.47]

= 4.77

pH + pOH = 14

pH = 14 - pOH

= 14 - 4.77

= 9.23

pH of the buffer = 9.23

queen_honey_blossom answered 4 months ago

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