How do you calculate PKa using the Henderson-Hasselbalch equation?

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How do you calculate PKa using the Henderson-Hasselbalch equation?

PH=6.98. Mix 25ml of 0.1M $N a H_{2} P O_{4}$ $NaH_2PO_4$ and 25ml of 0.1M $N a_{2} H P O_{4}$ $Na_2HPO_4$ and dilute with water in a 250ml Volumetric flask.

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Answer 1 · 2016-05-16T21:06:48

$p K_{a} = 6.98 - log (\frac{0.05 M}{0.05 M}) = 6.98$ $pK_a=6.98-log((0.05M)/(0.05M))=6.98$

Explanation:

The Henderson-Hasselbalch equation is:

$p H = p K_{a} + log (\frac{[B a s e]}{[A c i d]})$ $pH=pK_a+log(([Base])/([Acid]))$

Therefore, $p K_{a} = p H - log (\frac{[B a s e]}{[A c i d]})$ $pK_a=pH-log(([Base])/([Acid]))$

$p K_{a} = 6.98 - log (\frac{0.05 M}{0.05 M}) = 6.98$ $pK_a=6.98-log((0.05M)/(0.05M))=6.98$

Note that the new concentrations are divided by $2$ $2$ because the total volume doubles $V_{s o ln .} = 25 m L + 25 m L = 50 m L$ $V_(soln.)=25mL+25mL=50mL$

Notice that n this case the $p H = p K_{a}$ $pH=pK_a$ because the $[N a_{2} H P O_{4}] = [N a H_{2} P O_{4}]$ $[Na_2HPO_4]=[NaH_2PO_4]$ and the $log$ $log$ of the ratio is equal to zero.

Here is a video that explains more this topic (topic starts at minute 9:02):
Acid - Base Equilibria | Buffer Solution.

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How do you calculate PKa using the Henderson-Hasselbalch equation?

PH=6.98. Mix 25ml of 0.1M $N a H_{2} P O_{4}$ $NaH_2PO_4$ and 25ml of 0.1M $N a_{2} H P O_{4}$ $Na_2HPO_4$ and dilute with water in a 250ml Volumetric flask.

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Explanation:

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How do you calculate PKa using the Henderson-Hasselbalch equation?

PH=6.98. Mix 25ml of 0.1M NaH2PO4NaH_2PO_4 and 25ml of 0.1M Na2HPO4Na_2HPO_4 and dilute with water in a 250ml Volumetric flask.

1 Answer

Explanation:

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PH=6.98. Mix 25ml of 0.1M $N a H_{2} P O_{4}$ $NaH_2PO_4$ and 25ml of 0.1M $N a_{2} H P O_{4}$ $Na_2HPO_4$ and dilute with water in a 250ml Volumetric flask.