Question

Determine the [OH−] , pH, and pOH of a solution with a [H+] of 2.2×10−12 M at 25 °C. [OH−]= M pH= pOH= Determine the [H+] , pH, and pOH of a solution with an [OH−] of 9.7×10−5 M at 25 °C. [H+]= M pH= pOH= Determine the [H+] , [OH−] , and pOH of a solution with a pH of 10.30 at 25 °C. [H+]= M [OH−]= M pOH= Determine the [H+] , [OH−] , and pH of a solution with a pOH of 9.06 at 25 °C. [H+]= M [OH−]= M pH=

Answer 1

Answer:

For 1: The $OH^-$, pH and pOH of the solution is $4.57\times 10^{-3}M$, 11.66 and 2.36 respectively.

For 2: The $OH^-$, pH and pOH of the solution is $1.023\times 10^{-10}M$, 4.01 and 9.99 respectively.

For 3: The $H^+$, $OH^-$ and pOH of the solution is $5.012\times 10^{-11},1.99\times 10^{-5}M$ and 3.7 respectively.

For 4: The $H^+$, $OH^-$ and pOH of the solution is $8.71\times 10^{-10},1.07\times 10^{-5}M$ and 4.97 respectively.

Explanation:

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$       .....(1)

To calculate pOH of the solution, we use the equation:

$pH+pOH=14$       .....(2)

To calculate $[OH^]$ of the solution, we use the equation:

$pOH=-\log[OH^-]$     ......(3)

• For 1:

We are given:

$[H^+]=2.2\times 10^{-12}M$

Putting values in equation 1, we get:

$pH=-\log(2.2\times 10^{-12})\\\\pH=11.66$

Using equation 2, we get:

$pOH=14-11.66=2.34$

Now, using equation 3, we get:

$2.36=-\log[OH^-]$

$[OH^-]=10^{-2.36}=4.57\times 10^{-5}M$

Hence, the $OH^-$, pH and pOH of the solution is $4.57\times 10^{-3}M$, 11.66 and 2.36 respectively.

• For 2:

We are given:

$[H^+]=9.7\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(9.7\times 10^{-5})\\\\pH=4.01$

Using equation 2, we get:

$pOH=14-4.01=9.99$

Now, using equation 3, we get:

$9.99=-\log[OH^-]$

$[OH^-]=10^{-9.99}=1.023\times 10^{-10}M$

Hence, the $OH^-$, pH and pOH of the solution is $1.023\times 10^{-10}M$, 4.01 and 9.99 respectively.

• For 3:

We are given:

pH = 10.30

Putting values in equation 1, we get:

$10.30=-\log[H^+]$

$[H^+]=10^{-10.30)=5.012\times 10^{-11}M$

Using equation 2, we get:

$pOH=14-10.30=3.7$

Now, using equation 3, we get:

$3.7=-\log[OH^-]$

$[OH^-]=10^{-3.7}=1.99\times 10^{-4}M$

Hence, the $H^+$, $OH^-$ and pOH of the solution is $5.012\times 10^{-11},1.99\times 10^{-5}M$ and 3.7 respectively.

• For 4:

We are given:

pH = 9.06

Putting values in equation 1, we get:

$9.03=-\log[H^+]$

$[H^+]=10^{-9.30)=8.71\times 10^{-10}M$

Using equation 2, we get:

$pOH=14-9.03=4.97$

Now, using equation 3, we get:

$4.97=-\log[OH^-]$

$[OH^-]=10^{-4.97}=1.07\times 10^{-5}M$

Hence, the $H^+$, $OH^-$ and pOH of the solution is $8.71\times 10^{-10},1.07\times 10^{-5}M$ and 4.97 respectively.

Answer 2

pH is the negative logarithm of hydrogen ion concentration while pOH is the negative logarithm of hydroxide ion concentration.

a) We have that;

[H+] = 2.2×10−12 M

pH = - log  [H+]

pH =  - log [2.2×10−12 M]

pH = 11.65

Then

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 11.65

= 2.35

pOH = -log  [OH−]

[OH−] = Antilog -(pOH)

[OH−] = Antilog -(2.35)

[OH−] = 4.46 × 10^-3 M

b) [OH−] = 9.7×10−5 M

pOH =  -log  [OH−]

pOH =  -log[9.7×10−5 M]

pOH = 4.0

pH = 14 - 4.0

pH = 10

[H+] = Antilog (-10)

[H+] = 1 ×10−10

c) pH = 10.30

pOH = 14 - 10.30

pOH = 3.7

[OH−] = Antilog (-3.7)

[OH−] = 1.99 ×10−4

[H+] = -Antilog[10.30]

[H+] = 5.0 × 10−11

d) pOH =  9.06

pH = 14 - 9.06

pH = 4.94

[OH−] = Antilog(-9.06)

[OH−] = 8.7 × 10−10

pH = Antilog(-4.94)

pH = 1.1 × 10−5

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