Please help me answer this

Chemistry

1 answer:

Olenka [21]3 years ago

6 0

I think is the Third one tbh sorry if that’s not right

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If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?

Aleks [24]

pH = 2.1

Let $HA (aq)$ resembles the acid; as a weak acid (a small value of $K_{a}$ ) $HA$ would partially dissociate to produce protons $H^{+}$ and $A^{-}$ , its conjugate base. Let the final proton concentration (i.e., $[H^{+}]$ ) be $x$ . (Apparently $x \ge 0$ ) Construct the following RICE table:

$\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^{+}(aq) &+ &A^{-}(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right$

By definition, (all concentrations are under equilibrium condition)

$\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right$

It is given that

$K_a = 5.4 \cdot 10^{-6}$

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for $x$ gives:

$x^2 = 5.4 \cdot 10^{-6} \cdot (14 - x) \\x^2 + 5.4 \cdot 10^{-6} \cdot 14 \cdot x - 5.4 \cdot 10^{-6} \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)$