The $K_{a}$ $K_a$ for $H C O O H$ $HCOOH$ is $1.78 \cdot 10^{- 4}$ $1.78*10^-4$ . What would be the expected $[H_{3} O^{+}]$ $[H_3O^+]$ for a 1.0 M water solution of $H C O O H$ $HCOOH$ ?

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The $K_{a}$ $K_a$ for $H C O O H$ $HCOOH$ is $1.78 \cdot 10^{- 4}$ $1.78*10^-4$ . What would be the expected $[H_{3} O^{+}]$ $[H_3O^+]$ for a 1.0 M water solution of $H C O O H$ $HCOOH$ ?

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Answer 1 · 2018-04-30T04:52:01

The concentration of the hydronium ion, or $[H_{3} O^{+}]$ $[H_3O^+]$ , is $1.3 \cdot 10^{- 2} l M$ $1.3*10^-2color(white)(l)M$ .

Explanation:

For this, problem, we will use an ICE table.

To start, we need the net ionic equation for the reaction of $H C O O H$ $HCOOH$ , formic acid, with $H_{2} O$ $H_2O$ , water.

$H C O O H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H C O O^{-} (a q)$ $HCOOH(aq)+H_2O(l)rightleftharpoonsH_3O^"+"(aq)+HCOO^"-"(aq)$

Now, we can create our ICE table, where I stands for initial concentration, C stands for change, and E stands for equilibrium concentration.

$m m m H C O O H (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H C O O^{-} (a q)$ $color(white)(mmm)HCOOH(aq)+H_2O(l)rightleftharpoonsH_3O^+(aq)+HCOO^"-"(aq)$
$l$ $color(white)(l)$ -I- $m m 1.0 l M m m m m m - m m m l 0 l M m m m l m m 0 l M$ $color(white)(mm)1.0color(white)lMcolor(white)(mmmmm)-color(white)(mmml)0color(white)(l)Mcolor(white)(mmmlmm)0color(white)(l)M$

$l$ $color(white)(l)$ -C- $m m - x m m m m l m - m l m m + x l m m m m m + x$ $color(white)(mm)-xcolor(white)(mmmmlm)-color(white)(mlmm)+xcolor(white)(lmmmmm)+x$

$l$ $color(white)(l)$ -E- $l l m 1.0 - x m m m l m - m m m l m x m l m m m m l m x$ $color(white)(llm)1.0-xcolor(white)(mmmlm)-color(white)(mmmlm)xcolor(white)(mlmmmmlm)x$

$m$ $color(white)m$
Now, write the $K_{a}$ $K_a$ expression and plug the values in.

$m$ $color(white)m$
$K_{a} = \frac{[H_{3} O^{+}] [H C O O^{-}]}{[H C O O H]} = 1.78 \cdot 10^{- 4} = \frac{(x) (x)}{(1.0 - x)}$ $K_a=([H_3O^+][HCOO^-])/([HCOOH])=1.78*10^-4=((x)(x))/((1.0-x))$

Solve for $x$ $x$ . We will ignore the $x$ $x$ in the denominator, because $x$ $x$ will most likely be so small that it will be negligible when adding or subtracting.

$x^2/(1.0cancel(color(red)(-x)))=1.78*10^-4$

$x^2=1.78*10^-4$

$x=sqrt(1.78*10^-4)$

$x=1.3*10^-2$

$[H_3O^+]=1.3*10^-2color(white)(l)M$

We can now check if $x$ really was negligible when we ignored it before. This can be done by dividing $x$ by the equilibrium concentration of $HCOOH$ , multiplying by 100, and checking if the resultant value is less than $5%$ .

$(1.3*10^-2)/(1.0-1.3*10^-2)=1.32%<5%$

Because $1.32%$ is less than $5%$ , we can conclude that our choice to ignore $x$ earlier in the problem was valid.

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The $K_{a}$ $K_a$ for $H C O O H$ $HCOOH$ is $1.78 \cdot 10^{- 4}$ $1.78*10^-4$ . What would be the expected $[H_{3} O^{+}]$ $[H_3O^+]$ for a 1.0 M water solution of $H C O O H$ $HCOOH$ ?

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Explanation:

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The K_aKaK_a for HCOOHHCOOHHCOOH is 1.78*10^-41.78⋅10−41.78*10^-4. What would be the expected [H_3O^+][H3O+][H_3O^+] for a 1.0 M water solution of HCOOHHCOOHHCOOH?

1 Answer

Explanation:

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The $K_{a}$ $K_a$ for $H C O O H$ $HCOOH$ is $1.78 \cdot 10^{- 4}$ $1.78*10^-4$ . What would be the expected $[H_{3} O^{+}]$ $[H_3O^+]$ for a 1.0 M water solution of $H C O O H$ $HCOOH$ ?