If 30 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C? Ka for HC2H3O2 is 1.8 x 10^–5 at 25°C.

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If 30 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C? Ka for HC2H3O2 is 1.8 x 10^–5 at 25°C.

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Answer 1 · 2018-04-24T19:04:15

See below:

Explanation:

The reaction that will occur is:

$N a O H (a q) + C H_{3} C O O H (a q) \to C H_{3} C O O N a + H_{2} O (l)$ $NaOH(aq)+CH_3COOH(aq) -> CH_3COONa + H_2O(l)$

Now, using the concentration formula we can find the amount of moles of $N a O H$ $NaOH$ and Acetic acid:

$c = \frac{n}{v}$ $c=(n)/v$

For $N a O H$ $NaOH$
Remember that $v$ $v$ should be in litres, so divide any milliliter values by 1000.

$c v = n$ $cv=n$

$0.1 \times 0.03 = 0.003 m o l$ $0.1 times 0.03=0.003 mol$ of $N a O H$ $NaOH$

For $C H_{3} C O O H$ $CH_3COOH$ :

$c v = n$ $cv=n$

$0.2 \times 0.04 = 0.008 m o l$ $0.2 times 0.04=0.008 mol$ of $C H_{3} C O O H$ $CH_3COOH$ .

So 0.003 mol of $N a O H$ $NaOH$ will react to completion with the acid to form 0.003 mol of Sodium acetate, $C H_{3} C O O N a$ $CH_3COONa$ , in the solution, along with 0.005 mol of acid dissolved in a total volume of 70 ml. This will create an acidic buffer solution.

Let's find the concentration of the salt and the acid, respectively:

$c_{a c i d} = \frac{0.005}{0.7} \approx 0.0714 m o l d m^{- 3}$ $c_(acid)=(0.005)/0.7 approx 0.0714 mol dm^-3$

$c_{s a l t} = \frac{0.003}{0.007} \approx 0.0428 m o l d m^{- 3}$ $c_(sa l t)=(0.003)/0.007 approx 0.0428 mol dm^-3$

Now, we can use theHenderson-Hasselbalch equation to find the $p H$ $pH$ of the resulting solution.

The equation looks like this:

$p H = p K a + {log}_{10} (\frac{[S a l t]}{[A c i d]})$ $pH=pKa+log_10(([S a l t])/([Acid]))$

We are given the $K_{a}$ $K_a$ of the acid, so the $p K a$ $pKa$ is the negative logarithm of the $K_{a}$ $K_a$ value.

$p K a = - {log}_{10} [K_{a}]$ $pKa=-log_10[K_a]$
$p K a = - {log}_{10} [1.8 \times 10^{- 5}]$ $pKa=-log_10[1.8 times 10^-5]$
$p K a \approx 4.74$ $pKa approx 4.74$

Now we have to plug in all the values into the equation:

$p H = 4.74 + {log}_{10} (\frac{[0.0428]}{[0.0714]})$ $pH=4.74+log_10(([0.0428])/([0.0714]))$

$p H = 4.74 - 0.2218$ $pH=4.74-0.2218$

$p H \approx 4.52$ $pH approx 4.52$

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If 30 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C? Ka for HC2H3O2 is 1.8 x 10^–5 at 25°C.

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Explanation:

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