Chemistry, 26.12.2019 09:31 CrusaderLord

Calculate the ph for each of the following cases in the titration of 50.0 ml of 0.210 m hclo(aq) with 0.210 m koh(aq).
(a) before addition of any koh
(b) after addition of 25.0 ml of koh
(c) after addition of 35.0 ml of koh
(d) after addition of 50.0 ml of koh
(e) after addition of 60.0 ml of koh

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a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

Answer from: alwayspouty624

A measurement that tells about the acidic and basic nature of the given solution is called pH.

The equation for calculation:

$\text}{K}_{\text{a}} =\dfrac{ [H^{+}]^{2}}{\text{[HClO]}}$

Calculations:

(a) Before adding KOH:

$\text}{K}_{\text{a}} =\dfrac{ [H^{+}]^{2}}{\text{[HClO]}}$

Substituting values:

$4 \times 10^{-8} =\dfrac{ [H^{+}]^{2}}{\text{[0.21]}}$

$[H^{+}]^{2} =4 \times 10^{-8} \times \text{0.21}$

$[H^{+}]^{2} &= 8.4 \times 10^{-9 }$

Now solving further without square:

$[H^{+}] &= \sqrt{8.4 \times 10^{-9 }}$

$[H^{+}] = 9.2 \times 10^{-5} \text{M}$

Calculating pH:

$\text{pH} = - \text{log} [H^{+}]$

$\text{pH} = - \text{log} (9.2 \times 10^{-5})$

pH = 4

(b) After adding 25 mL KOH:

This will cause the solution to act as a buffer and hence Henderson-Hasselbalch equation will be used.

$\rm pH = Pka + log\dfrac{[Salt]}{[Acid]}$

Moles (n) of HCIO= Molarity (M) × Volume (V)

Moles = 0.21M × 0.05L

n = 0.0105 moles

Moles (n) of KOH = 0.21 × 0.025

n =0.00525 moles

Therefore, remaining moles = 0.0105 - 0.00525

Remaining moles = 0.00525

Total volume (V) is = 0.05 L + 0.025 L

V = 0.075 L

$\rm Molarity (M) of\; HCIO = \dfrac {Moles \;HCIO\; remaining } {Total\; volume}$

$\rm M = \dfrac {0.00525 }{0.075}$

M = 0.07 M

$\rm Molarity (M) of\; KCIO = \dfrac {Moles \;of \;KCIO\;} {Total\; volume}$

$\rm M = \dfrac {0.00525 }{0.075}$

M = 0.07 M

Given, Ka = 4 x 10⁻⁸

Since, pKa = -㏒ Ka

$\rm pKa = - log (4 \times 10^{-8})$

pKa = 7.4

By substituting values in equation:

$\rm pH = 7.4 +log (\dfrac{0.07}{0.07})$

∴ pH = 7.4

This is again a buffer solution and we will use Henderson-Hasselbalch equation.

Moles (n) of HCIO = Molarity (M) × Volume (V)

n = 0.21 M × 0.05L

n = 0.0105 moles

Moles (n) of KOH = Molarity (M) × Volume (V)

n = 0.22 M × 0.035 L

n = 0.0077 moles

Remaining moles = 0.0105 - 0.0077

Remaining moles = 0.0028

Total volume (V) = 0.05L + 0.035L V

V = 0.085 L

$\rm Molarity (M) of\; HCIO = \dfrac {Moles \;HCIO\; remaining } {Total\; volume}$

$\rm M = \dfrac{8 \times 10^{-5}}{ 0.085}$

M = 9.4 x 10⁻⁴ M

$\rm Molarity (M) of\; KCIO = \dfrac {Moles \;of \;KCIO\;} {Total\; volume}$

$\rm M = \dfrac {0.0077M }{0.085 L}$

M = 0.09 M

Substituting value we get:

$\rm pH = 7.4 + log ( \dfrac{0.09 }{9.4 \times 10^{-4}})$

Therefore, pH = 8.38

(d) After adding 50 mL:

$\rm Molarity (M) of\; KCIO = \dfrac {Moles \;of \;KCIO\;} {Total\; volume}$

$\rm M = \dfrac{0.0105 \;mol} {0.1 L}$

M = 0.105 M

$\rm Ka = \dfrac {KW } {Kb}$

$\rm Ka = \dfrac {1 x 10^{-14}} {4 x 10^{-8}}$

Ka = 2.5 x 10⁻⁷

Expression of Kb will be used as:

$\rm Kb = \dfrac {[CIO^{-}] [OH^{-}] }{ [KCIO]}$

In this [CIO⁻] = [OH⁻] so,

$\rm Kb = \dfrac {[OH^{-}] ^{2}}{ [KCIO]}$

$\rm 2.5 x 10^{-7}= \dfrac {[OH^{-}] ^{2}}{ [0.105]}$

Since, [OH⁻] = 0.00016 M

$\rm pOH = - log [OH^{-}]$

pOH = - log 0.00016

pOH = 3.8

Since, pH = 14 - pOH

pH = 14 - 3.8

pH = 10.2

(e) After adding 60 mL KOH:

Molarity of KOH can be calculated by:

M₁ × V₁ = M₂ × V₂

Given,

M₁ = ?V₁ = 0.05 + 0.06 V₁ = 0.11 LM₂ = 0.21 M V₂ = 0.01L

By placing values in the equation :

M₁ × 0.11 L = 0.21 × 0.01 L

M₁ =0.02 M

Since, [KOH] = [OH⁻] = 0.02 M

$\rm pOH = - log [OH^{-}]$

pOH = - log 0.02

pOH = 1.7

Calculating for pH:

pH = 14 - pOH

pH= 14 - 1.7

pH = 12.3

Therefore, pH are 4, 7.4, 8.38, 10.2 and 12.3 .

To learn more about pH and Henderson-Hasselbalch equation follow the link:

link

Answer from: Quest

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

Answer from: Quest

c. 75.7

i think that is the right

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Calculate the ph for each of the following cases in the titration of 50.0 ml of 0.210 m hclo(aq) wit...

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