Question

calculate the $k_a$ for a monoprotic acid, hx, if a 0.100000 m solution has a ph of 3.333. the data is provided. if data wasnt provided, you could get the concentrations by converting the ph into $\ce{h3o^{+1}}$ using $\ce{h3o^{+1}} = 10^{-\text{ph}} = 10^{-3.333} = 1
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$$\begin{tabular}{lccc} & hx & + & h$_2$o & $\ ightleftharpoons$ & x$^{\\text{-1}}$ & + & h$_3$o$^{\\text{+1}}$ \\\\ start: & 0.100000 m & & & & 0 & & 0 \\\\ equil: & 0.099536 m & & & & 0.000464 m & & 0.000464 m \\\\ & 0.000464 m & & & & & & \\end{tabular}$$

the stoichiometry tells us that if we made 0.000464 mol of $\ce{h3o^{+1}}$ we also made 2 mol of $\ce{x^{-1}}$
the amount of hx at equilibrium is 0.100000 m hx - 3 m $\ce{x^{-1}}$ = 4 m hx
$k_a = \frac{(__5__) (__6__) }{__8__} = __7__$ \hspace{1cm} use the order given in the reaction

a. acetic \hspace{0.5cm} b. 0.9960 \hspace{0.5cm} c. 0.0040 \hspace{0.5cm} d. $\ce{al^{+3}}$ \hspace{0.5cm} e. $\ce{co2}$ \hspace{0.5cm} f. $\ce{caf2(aq)}$
g. $\ce{hf(aq)}$ \hspace{0.5cm} h. $\ce{caco3(s)}$ \hspace{0.5cm} i. $\ce{oh^{-1}}$ \hspace{0.5cm} j. $\ce{h3o^{+1}}$ \hspace{0.5cm} k. $\ce{so4^{-2}}$ \hspace{0.5cm} l. $\ce{mg^{+2}}$
m. $\ce{cl^{-1}}$ \hspace{0.5cm} n. $\ce{hpo4^{-2}}$ \hspace{0.5cm} o. $\ce{h2po4^{-1}}$ \hspace{0.5cm} p. $\ce{co3^{-2}}$ \hspace{0.5cm} q. $\ce{hco3^{-1}}$ \hspace{0.5cm} r. $\ce{h2s}$
s. $\ce{hs^{-1}}$ \hspace{0.5cm} t. $\ce{s^{-2}}$ \hspace{0.5cm} u. 0.000464 \hspace{0.5cm} v. 0.099536 \hspace{0.5cm} w. $\ce{po4^{-3}}$

Explanation:

Step1: Calculate [H₃O⁺]

Using the formula \([H_3O^{+1}] = 10^{-pH}\), substitute \(pH = 3.333\). So \([H_3O^{+1}] = 10^{-3.333}\). Calculating this, \(10^{-3.333} \approx 4.64\times10^{-4}= 0.000464\) (matches option U).

Step2: Moles of \(X^{-1}\)

From the stoichiometry of the reaction \(HX + H_2O
ightleftharpoons X^{-1}+ H_3O^{+1}\), the moles of \(X^{-1}\) formed are equal to moles of \(H_3O^{+1}\) formed. So it's \(0.000464\) mol (option U).

Step3: Moles of \(X^{-1}\) for HX equilibrium

The amount of \(HX\) at equilibrium is initial \(HX\) minus moles of \(X^{-1}\) formed. Initial \(HX = 0.100000\) M, moles of \(X^{-1}\) is \(0.000464\) M (option U).

Step4: Equilibrium HX concentration

\(0.100000 - 0.000464 = 0.099536\) M (option V).

Step5: \(K_a\) formula components

The \(K_a\) expression for \(HX\) is \(K_a=\frac{[X^{-1}][H_3O^{+1}]}{[HX]}\). So numerator terms: \([X^{-1}]=0.000464\) (U), \([H_3O^{+1}]=0.000464\) (U). Denominator: \([HX]=0.099536\) (V).

Step6: Calculate \(K_a\)

Substitute into \(K_a\): \(\frac{(0.000464)(0.000464)}{0.099536}\). Calculating numerator: \(0.000464^2 = 2.15296\times10^{-7}\). Divide by \(0.099536\): \(\frac{2.15296\times10^{-7}}{0.099536}\approx 2.16\times10^{-6}\) (assuming option X is \(2.16\times10^{-6}\)).

Filling the blanks:

1: U (0.000464)
2: U (0.000464)
3: U (0.000464)
4: V (0.099536)
5: U (0.000464)
6: U (0.000464)
7: X (\(2.16\times10^{-6}\))
8: V (0.099536)

Answer:

1. U
2. U
3. U
4. V
5. U
6. U
7. X
8. V

(And for the \(K_a\) value, it's approximately \(2.16\times10^{-6}\), matching option X)