Question

Calculate the number of moles of NH4Cl that must be added to 2.0 L of 0.10 M NH3 to

form a buffer with a pH of 9.00 given that pKb = 9.256.

(Assume that the addition of NH4Cl does not change the volume of the solution). ​

Answer 1

Explanation:

I don't know the answer sorry

Answer 2

The no. of moles of NH₄Cl to be added to the given NH₃ solution to form the buffer of pH 9.00 is $1.109$×$10^{-5}$

Explanation:

For a basic solution,

$pOH = -log Kb + log\frac{[salt]}{[base]}$

$pOH = 14 - pH \\ = 14 - 9 = 5$

$pKb = -logKb$

Given that molarity of NH₃ is $0.10$ M

Therefore,

$5 = 9.256 + log\frac{[salt]}{0.1}$

$-4.256 = log\frac{[salt]}{0.1}$

Taking antilog both sides

we get,

$5.546$×$10^{-5}$ $= \frac{[salt]}{0.1}$

$[salt] = 5.546$×$10^{-6}$

Moles of NH₄Cl = Molarity of NH₄Cl × Volume of solution (2.0 L)

$= 5.546$×$10^{-6}$ × $2$

$= 1.109$×$10^{-5}$