Question

50 POINTS PLEASE HELP ASAP! I WILL MARK BRAINLIEST ONLY IF CORRECT AND CLEARLY STATED

19. Calculate the hydrogen-ion concentration [H^+] for the aqueous solution in which [OH^-] is 1 x 10^-11 mol/L. Is this solution acidic, basic, or neutral? Show your work. (3 points)

20. Calculate the acid dissociation constant of a weak monoprotic acid if a 0.5 M solution of this acid gives a hydrogen-ion concentration of 0.000 1.M? Show your work. (3 points)

DO NOT Copy and past the work from another answer UNLESS you can clearly state all the steps. I've seen them all so I know If you copy.

Answer 1

Answer:

1) The hydrogen-ion concentration is 0.001 M and pH of the solution is less than 7 which means that the solution is acidic.

2) The acid dissociation constant of a weak monoprotic acid is $2.0\times 10^{-8}$.

Explanation:

1) Concentration of hydroxide ions =$[OH^-]=1\times 10^{-11}$

The pOH of the solution:

$pOH=-\log[OH^-]$

$pOH=-\log[1\times 10^{-11}=11$

pH + pOH = 14

pH = 14 - pOH = 14 - 11 = 3

The pH of the solution is given as:

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

$3=-\log[H^+]$

$[H^+]=10^{-3} M=0.001 M$

The hydrogen-ion concentration is 0.001 M and pH of the solution is less than 7 which means that the solution is acidic.

2)

Dissociation on weak monoprotic weak acid is given by :

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

Initially

0.5 M             0   0

At equilibrium

(0.5-x)            x    x

Given , hydrogen ions concentration = $[H^+]=x=0.0001 M$

The expression of a dissociation constant$K_a$  is given as:

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

$=\frac{x\times x}{(0.5-x)}=\frac{0.0001\times 0.0001 }{(0.5-0.0001)}$

$K_a=2.0\times 10^{-8}$

The acid dissociation constant of a weak monoprotic acid is $2.0\times 10^{-8}$.

Answer 2

Answer:

1) The hydrogen-ion concentration is 0.001 M and pH of the solution is less than 7 which means that the solution is acidic.

2) The acid dissociation constant of a weak monoprotic acid is 2.0\times 10^{-8}2.0×10

−8

.

Explanation:

1) Concentration of hydroxide ions =[OH^-]=1\times 10^{-11}[OH

−

]=1×10

−11

The pOH of the solution:

pOH=-\log[OH^-]pOH=−log[OH

−

]

pOH=-\log[1\times 10^{-11}=11pOH=−log[1×10

−11

=11

pH + pOH = 14

pH = 14 - pOH = 14 - 11 = 3

The pH of the solution is given as:

pH=-\log[H^+]pH=−log[H

+

]

3=-\log[H^+]3=−log[H

+

]

[H^+]=10^{-3} M=0.001 M[H

+

]=10

−3

M=0.001M

The hydrogen-ion concentration is 0.001 M and pH of the solution is less than 7 which means that the solution is acidic.

2)

Dissociation on weak monoprotic weak acid is given by :

HAleftharpoons A^-+H^+HAleftharpoonsA

−

+H

+

Initially

0.5 M 0 0

At equilibrium

(0.5-x) x x

Given , hydrogen ions concentration = [H^+]=x=0.0001 M[H

+

]=x=0.0001M

The expression of a dissociation constantK_aK

a

is given as:

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

a

=

[HA]

[A

−

][H

+

]

=\frac{x\times x}{(0.5-x)}=\frac{0.0001\times 0.0001 }{(0.5-0.0001)}=

(0.5−x)

x×x

=

(0.5−0.0001)

0.0001×0.0001

K_a=2.0\times 10^{-8}K

a

=2.0×10

−8

The acid dissociation constant of a weak monoprotic acid is 2.0\times 10^{-8}2.0×10

−8

Explanation:

hope it helps you

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