Question

question 4 (1 point)
write the expression for the $k_a$ when $hc_2h_3o_2$ is dissolved in water
$hc_2h_3o_2 + h_2o <-> c_2h_3o_2^{-1} + h_3o^{+1}$
use the order given in the reaction above
\\ k_a = \frac{\\_\\_1\\_\\_ \\_\\_2\\_\\_}{\\_\\_3\\_\\_} \\
a. $h_20_{(liq)}$ b. $oh^{-1}$ c. $h_3o^{+1}$ d. $so_4^{-2}$ e. $mg^{+2}$ f. $cl^{-1}$
g. $hpo_4^{-2}$ h. $h_2po_4^{-1}$ i. $co_3^{-2}$ j. $hco_3^{-1}$ k. $h_2s$ l. $hs^{-1}$
m. $s^{-2}$ n. $h_2o$ o. $h_2co_3$ p. $h_3po_4$ q. $hc_2h_3o_2$ r. $c_2h_3o_2^{-1}$
s. $po_4^{-3}$ t. $so_4^{-2}$ u. $caco_3{(s)}$ v. $hf_{(aq)}$

Explanation:

Step1: Recall \( K_a \) expression rules

For a weak acid dissociation \( HA + H_2O
ightleftharpoons A^- + H_3O^+ \), the acid dissociation constant \( K_a \) is given by \( K_a=\frac{[A^-][H_3O^+]}{[HA]} \), where pure liquids (like \( H_2O_{(l)} \)) are not included in the equilibrium expression.

Step2: Identify species in the reaction

The reaction is \( HC_2H_3O_2 + H_2O
ightleftharpoons C_2H_3O_2^{-1} + H_3O^{+1} \). Here, \( HA = HC_2H_3O_2 \) (Q), \( A^- = C_2H_3O_2^{-1} \) (R), and \( H_3O^+ = H_3O^{+1} \) (C). \( H_2O \) is a liquid, so it's excluded.

Step3: Construct \( K_a \) expression

Substituting into the \( K_a \) formula: \( K_a=\frac{[C_2H_3O_2^{-1}][H_3O^{+1}]}{[HC_2H_3O_2]} \), which corresponds to [R] [C] in the numerator and [Q] in the denominator. But looking at the options, the numerator first blank (1) is \( C_2H_3O_2^{-1} \) (R), second blank (2) is \( H_3O^{+1} \) (C), and denominator blank (3) is \( HC_2H_3O_2 \) (Q). Wait, but the options for the blanks: the first numerator blank (1) is [1] which is \( C_2H_3O_2^{-1} \) (R), second (2) is \( H_3O^{+1} \) (C), and denominator (3) is \( HC_2H_3O_2 \) (Q). But let's check the option letters: R is \( C_2H_3O_2^{-1} \), C is \( H_3O^{+1} \), Q is \( HC_2H_3O_2 \). Wait, maybe the labels: the first numerator term (1) is R, second (2) is C, denominator (3) is Q. But let's confirm the \( K_a \) expression structure. The reaction is \( HC_2H_3O_2(aq) + H_2O(l)
ightleftharpoons C_2H_3O_2^-(aq) + H_3O^+(aq) \). So \( K_a = \frac{[C_2H_3O_2^-][H_3O^+]}{[HC_2H_3O_2]} \). So blank 1: \( C_2H_3O_2^- \) (R), blank 2: \( H_3O^+ \) (C), blank 3: \( HC_2H_3O_2 \) (Q). But the options for the blanks (the dropdowns) have options, but the question is about the letters. Wait, the first numerator box (1) should be R ( \( C_2H_3O_2^{-1} \) ), second (2) is C ( \( H_3O^{+1} \) ), and denominator (3) is Q ( \( HC_2H_3O_2 \) ). But let's check the option list: R is \( C_2H_3O_2^{-1} \), C is \( H_3O^{+1} \), Q is \( HC_2H_3O_2 \). So the expression is \( K_a = \frac{[R][C]}{[Q]} \), but in the given format, the numerator is [1] [2] and denominator [3]. So 1: R, 2: C, 3: Q. But let's check the options again. Wait, the options for the first blank (1) are the choices, so the first term in numerator is \( C_2H_3O_2^{-1} \) (R), second is \( H_3O^{+1} \) (C), denominator is \( HC_2H_3O_2 \) (Q). So the answer for the blanks: 1: R, 2: C, 3: Q. But the question's options for the blanks (the dropdowns) have the options as per the list. So the \( K_a \) expression is \( \frac{[C_2H_3O_2^{-1}][H_3O^{+1}]}{[HC_2H_3O_2]} \), so [R] [C] over [Q]. But the problem's format is \( K_a = \frac{[__1__][__2__]}{[__3__]} \), so 1 is R ( \( C_2H_3O_2^{-1} \) ), 2 is C ( \( H_3O^{+1} \) ), 3 is Q ( \( HC_2H_3O_2 \) ).

Answer:

For the numerator first blank (1): R. \( C_2H_3O_2^{-1} \)
For the numerator second blank (2): C. \( H_3O^{+1} \)
For the denominator blank (3): Q. \( HC_2H_3O_2 \)

(In the format of the question's blanks: \( K_a = \frac{[R][C]}{[Q]} \))