Question

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

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}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

❤ ❤ ❤ ❤ ❤

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 $2.0\times 10^{-8}$.

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 $2.0\times 10^{-8}$.