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Setler79 [48]
2 years ago
10

When a 0. 1 m aqueous solution of hydrocyanic acid, hcn, reaches equilibrium, the ph is measured to be 5. 20. using this informa

tion, calculate ka for hydrocyanic acid
Chemistry
1 answer:
Sav [38]2 years ago
7 0

When a 0. 1 m aqueous solution of hydrocyanic acid, HCN, reaches equilibrium, the ka for hydrocyanic acid is  3.969 x 10⁻¹⁰.

<h3>What is ka value?</h3>

It's the value of equilibrium constant for the dissociation of ions into a solution. The more the Ka value the more will be dissociation.

Ka = [H₃O⁺]² / [HCN] [H₃O⁺]

The pH is 5.20

-log [H₃O⁺] = 5.20

Putting antitlog both side.

The value will be 6.30 x 10⁻⁶

Ka = (6.30 x 10⁻⁶)²  / 0.1 - 6.30 x 10⁻⁶

0.1 - 6.30 x 10⁻⁶ = 0.1

Ka = 3.969 x 10⁻¹⁰

Thus, the Ka value for hydrocyanic acid is 3.969 x 10⁻¹⁰.

To learn more about ka value, refer to the link:

brainly.com/question/2796803

#SPJ4

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Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
A buffer solution is composed of 4.00 4.00 mol of acid and 3.25 3.25 mol of the conjugate base. If the p K a pKa of the acid is
Reika [66]

<u>Answer:</u> The pH of the buffer is 4.61

<u>Explanation:</u>

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of weak acid = 4.70

[\text{conjuagate base}]} = moles of conjugate base = 3.25 moles

[\text{acid}]  = Moles of acid = 4.00 moles

pH = ?

Putting values in above equation, we get:

pH=4.70+\log(\frac{3.25}{4.00})\\\\pH=4.61

Hence, the pH of the buffer is 4.61

8 0
3 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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