What is the pH of 0.256 M ammonium chloride NH4Cl?

Calculate pH of a buffer solution with 1.0 M ammonia (NH3 ; Kb = 1.8 x 10-5) & 1.0 M ammonium chloride (NH4Cl)?

Answer:

For this, we can use the Henderson-Hasselbalch Equation, which states that: pH = pKa + log([base]/[acid]) We can find the pKa of the solution by finding Ka from Kb Kb*Ka = 10^-14 Ka = (10^-14)/(1.8*10^-5) Ka = 5.56*10^-10 pKa = -logKa pKa = 9.2553 Now, we can say: pH = 9.2553 + log([base]/[acid]) We have 1 M ammonia, the base, and 1 M ammonium ions, the acid, so now we have: pH = 9.2553 + log(1/1) pH = 9.2553 + 0 pH = 9.2553

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For this, we can use the Henderson-Hasselbalch Equation, which states that: pH = pKa + log([base]/[acid]) We can find the pKa of the solution by finding Ka from Kb Kb*Ka = 10^-14 Ka = (10^-14)/(1.8*10^-5) Ka = 5.56*10^-10 pKa = -logKa pKa = 9.2553 Now, we can say: pH = 9.2553 + log([base]/[acid]) We have 1 M ammonia, the base, and 1 M ammonium ions, the acid, so now we have: pH = 9.2553 + log(1/1) pH = 9.2553 + 0 pH = 9.2553

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