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Example 2:

Calculate the pH of a 0.30M NH3/0.36M NH4Cl buffer system (Ka= 5.6 x10-10).

asked Mar 18, 2015 in CHEMISTRY by heather Apprentice

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1 Answer

Step 1:

The Concentration of $\small NH_{3}$ is 0.30 M.

The Concentration of $\small \fn_cm \small NH_{4}Cl$ is 0.36 M.

Equilibrium constant $\small k_{a} =5.6 \times 10^{-10}$ .

pH definition:

It is the measure of the acidity and alkalinity of a solution.

pH is the negative log of the activity of the hydrogen ion in an aqueous solution.

$\small pH = -log\left ( k_{a} \right ) +log\left (\frac{[A^{-}]}{[HA]} \right )$

$\small pH = -log(5.6\times 10^{-10})+log\left (\frac{NH_{3}}{NH_{4}Cl} \right )$

$\small pH = -log(5.6\times 10^{-10})+log\left (\frac{0.3}{0.36} \right )$

$\small \\pH =9.2518-0.0791 \\\\pH = 9.1726$

Solution:

pH of the buffer system is 9.17.

answered Mar 18, 2015 by Lucy Mentor

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