Question

How many grams of dry NH4Cl need to be added to 1.90 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.72? Kb for ammonia is 1.8×10−5.

Answer 1

Answer : The mass of dry $NH_4Cl$ needed are, 174.196 grams.

Explanation :

First we have to calculate the pOH of the solution.

$pH+pOH=14$

$pOH=14-pOH\\\\pOH=14-8.72\\\\pOH=5.28$

Now we have to calculate the $pK_b$.

$pK_b=-\log K_b$

$pK_b=-\log (1.8\times 10^{-5})$

$pK_b=4.745$

Now we have to calculate the concentration of base, $NH_4Cl$.

Using Henderson Hesselbach equation :

$pOH=pKb+\log \frac{[Salt]}{[Base]}$

$pOH=pKb+\log \frac{[NH_4Cl]}{[NH_3]}$

Now put all the given values in this equation, we get :

$5.28=4.745+\log \frac{[NH_4Cl]}{0.500}$

$[NH_4Cl]=1.714M$

Now we have to calculate the mass of $NH_4Cl$.

Formula used :

$Molarity=\frac{\text{Moles of }NH_4Cl}{\text{Volume of solution}}$

$Molarity=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl\times \text{Volume of solution}}$

Now put all the given values in this formula, we get:

$1.714M=\frac{\text{Mass of }NH_4Cl}{53.49\times 1.90L}$

$\text{Mass of }NH_4Cl=174.196g$

Therefore, the mass of dry $NH_4Cl$ needed are, 174.196 grams.

Answer 2

The pH has been the hydrogen ion concentration and pOH is the hydroxide ion concentration. The mass of ammonium chloride in the solution is 174.196 grams.

What is Kb?

The Kb is the base dissociation constant for the compound.

The pH of the solution is 8.72, the pOH of the solution is given as:

$\rm pH=14-pOH\\8.72=14-pOH\\pOH=5.28$

The pOH of the given solution is 5.28. The concentration of ammonium chloride salt in the solution from pOH can be given as:

$\rm pOH=pKb\;+\;log\dfrac{salt}{acid}$

The pKb has been given as the logarithmic value of Kb. The concentration of ammonia is 0.5 M. Substituting the values for the concentration of ammonium chloride salt as:

$\rm 5.28=log\;1.8\;\times\;10^{-5}\;+\;log\dfrac{NH_4Cl}{0.5}\\\\ 5.28=4.745\;+\;log\dfrac{NH_4Cl}{0.5}\\\\NH_4Cl=1.714\;M$

The concentration of ammonium chloride salt is 1.714 M. The volume of the solution is 1.90 L. The molar mass of the compound is 53.49 g/mol.

Substituting the values for the mass of ammonium chloride as:

$\rm Molarity=\dfrac{mass}{molar\;mass\;\times\;volume} \\\\1.714\;M=\dfrac{mass}{53.49\;g/mol\;\times\;1.90\;L}\\\\ Mass=174.196\;g$

The mass of ammonium chloride added to the solution is 174.196 grams.

Learn more about base dissociation constant, here:

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