Question

Calculate the pH of a buffer solution created by reacting 100 mL of 0.1 M NH3 with 90 mL of 0.1 M HNO3. (Remember, you can find Ka and Kb values on gchem!)

Answer 1

This question is asking for the pH of a buffer solution between ammonia and nitric acid, with given volumes and concentrations. At the end, the result turns out to be 10.488.

Buffers

In chemistry, buffers are known as substances attempting to hold a relatively constant pH by mixing and acid and a base (weak and strong). In such a way, for the substances given, the first step will be to calculate the consumed moles as they are mixed:

$n_{NH_3}=0.1L*0.1mol/L=0.01mol\\\\n_{HNO_3}=0.09L*0.1mol/L=0.009mol$

Now, since ammonia is in a greater proportion, one can calculate how much of it is left after being consumed by the nitric acid:

$n_{NH_3}^{left}=0.01mol-0.009mol=0.001mol$

And its new concentration:

$[NH_3]=\frac{0.001mol}{0.1L+0.09L} =0.00526M$

Next, with ammonia's ionization:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We set up the equilibrium expression based on ammonia's Kb:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Which can be solved by introducing x and using ammonia's Kb:

$1.8x10^{-5}=\frac{x^2}{0.00526M}$

Then, we solve for x which is also equal to the concentration of ammonium and hydroxide ions in the solution:

$x=\sqrt{0.00526*1.8x10^{-5}}=0.000308M$

Ultimately, we calculate the pOH and then turn it into pH with:

$pOH=-log(0.00308)=3.512\\\\pH=14-3.512=10.488$

Learn more about buffers: brainly.com/question/24188850