Question

What is the ph of a buffer system prepared by dissolving 10.70 grams of nh4cl and 35.00 ml of 12 m nh3 in enough water to make 1.000 l of solution? kb = 1.80 × 10-5 for nh3?

Answer 1

Answer:- 9.58

Solution:- We can calculate the pH for the given one using Handerson equation.

pH = pKa + log(base/acid)

Ammonia is the base and ammonium ion (ammonium chloride) is the acid.

We have 10.70 grams of ammonium chloride. Molar mass of ammonium chloride is 53.49 g/mol.

moles of ammonium chloride or ammonium ion = 10.70 g x (1mol/53.49 g)

= 0.2000 mol

We have 35.00 ml of 12 M ammonia solution. So, we can calculate the moles of ammonia also.

35.00 ml is 0.0350 L, So....

0.0350 L x (12 mol/1L) = 0.420 mol

Volume of the buffer is 1.00 liter, so the molarities for both the solutions would be same as their number of moles.

[base] = 0.420 M and [acid] = 0.200 M

We can calculate pKb from given kb value.

pKb = - log Kb

pKb = $- log(1.80*10^-5)$

pKb = 4.74

Now we can calculate pKa as...

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

So, pH = $9.26 + log(\frac{0.420}{0.200})$

pH = 9.26 + 0.322

pH = 9.58