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Alex_Xolod [135]
3 years ago
15

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places

Chemistry
2 answers:
Lemur [1.5K]3 years ago
8 0

The pH of a 0.20 M solution of KCN is \boxed{11.31}.

Further Explanation:

pH is used to describe acidity or basicity of substances. Its range varies from 0 to 14. It is defined as negative logarithmof hydrogen ion concentration.

The expression for pH is mentioned below.

{\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right]                                                                     …… (1)

Where \left[ {{{\text{H}}^ + }}\right] is the concentration of hydrogen ion.

Dissociation reaction of KCN is as follows:

{\text{KCN}} \to {{\text{K}}^ + } + {\text{C}}{{\text{N}}^ - }  

Cyanide ions thus formed can react with water to form HCN and {\text{O}}{{\text{H}}^ - } as follows:

{\text{C}}{{\text{N}}^ - } + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{HCN}} + {\text{O}}{{\text{H}}^ - }  

The relation between {{\text{K}}_{\text{w}}}, {{\text{K}}_{\text{b}}} and {{\text{K}}_{\text{a}}} is expressed by following relation:

{{\text{K}}_{\text{w}}} = {{\text{K}}_{\text{b}}} \cdot {{\text{K}}_{\text{a}}}                                                                                 …… (2)

Where,

{{\text{K}}_{\text{w}}} is the ionic product constant of water.

{{\text{K}}_{\text{b}}} is the dissociation constant of base.

{{\text{K}}_{\text{a}}} is the dissociation constant of acid.

The value of {{\text{K}}_{\text{w}}} is {10^{ - 14}}.

The value of {{\text{K}}_{\text{a}}} is 4.9 \times {10^{ - 10}}.

Substitute these values in equation (2).

{10^{ - 14}} = {{\text{K}}_{\text{b}}}\left( {4.9 \times {{10}^{ - 10}}} \right)  

Solve for {{\text{K}}_{\text{b}}},

{{\text{K}}_{\text{b}}} = 2 \times {10^{ - 5}}  

The expression for {{\text{K}}_{\text{b}}} of HCN is as follows:

{{\text{K}}_{\text{b}}} = \dfrac{{\left[ {{\text{HCN}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{C}}{{\text{N}}^ - }} \right]}}                                                                            …… (3)

Consider x to be change in equilibrium concentration. Therefore, equilibrium concentrationof {\text{C}}{{\text{N}}^ - }, HCN and   becomes (0.2 – x), x and x respectively.

{\text{2}} \times {\text{1}}{{\text{0}}^{ - 5}} = \dfrac{{{x^2}}}{{\left( {0.2 - x} \right)}}  

Solving for x,

x = 0.002  

Therefore concentration of hydroxide ion is 0.002 M.

The expression to calculate pOH is as follows:

{\text{pOH}} =  - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]                                                                             …… (4)

Substitute 0.002 M for \left[ {{\text{O}}{{\text{H}}^ - }} \right] in equation (4).

\begin{aligned}{\text{pOH}} &= - \log \left( {0.002{\text{ M}}} \right) \\&= 2.69 \\\end{aligned}  

The relation between pH and pOH is as follows:

pH + pOH = 14                                                                          …… (5)

Substitute 2.69 for pOH in equation (4).

{\text{pH}} + 2.69 = 14  

Solving for pH,

pH = 11.31

Learn more:

  1. Write the chemical equation responsible for pH of buffer containing  and  : brainly.com/question/8851686
  2. Reason for the acidic and basic nature of amino acid. brainly.com/question/5050077

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acids, base and salts

Keywords: pH, pOH, 11.31, 2.69, 14, 0.002 M, Kb, Kw, Ka, 10^-14, 2*10^-5.

storchak [24]3 years ago
4 0
The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

          CN-    +    H2O    ⇔    HCN +    OH-

I      0.2 m             ∞                   0             0
C      -x                 ∞                   +x        +x
-----------------------------------------------------------
E      0.2-x                               +x         +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

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Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

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Answer: 19.25 gallons

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Given:  4 quarts = 1 gallon

Thus if 1 ml is equal to 0.0011 quart

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Now if 4 quarts is equal to 1 gallon.

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