2 years ago

PLS HELP ITS URGENT

Chemistry

1 answer:

mamaluj [8]2 years ago

6 0

Answer:

Answer

The answer is option 4

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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0144 0.0144 M solution. The pH of the resulting solution i

alexdok [17]

Answer:

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

Explanation:

The pH of the solution = 2.46

$pH=-\log[H^+]$

$2.46=-\log[H^+]$

$[H^+]=0.003467 M$

$HA\rightleftharpoons H^++A^-$

Initially

0.0144 0 0

At equilibrium

(0.0144-x) x x

The expression if an dissociation constant is given by :

$K_a=\frac{[A^-][H^+]}{[HA]}$

$K_a=\frac{x\times x}{(0.0144-x)}$

$x=[H^+]=0.003467 M$

$K_a=\frac{0.003467 \times 0.003467 }{(0.0144-0.003467 )}$

$K_a=1.099\times 10^{-3}$

The value of dissociation constant of the monoprotic acid is $1.099\times 10^{-3}$ .

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3 years ago

Can someone please help

miskamm [114]

Answer:

Explanation:

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