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BigorU [14]
3 years ago
9

If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?

Chemistry
1 answer:
valkas [14]3 years ago
6 0

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

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Explanation: hope this helped

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Living things respire to stay alive. In the mitochondria, sugar is combined with oxygen to produce energy, water and carbon diox
olya-2409 [2.1K]

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C6H12O6 + 6O2 → 6CO2 + 6H2O

Explanation:

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8 0
2 years ago
Read 2 more answers
6. The given mass of helium occupies
marin [14]

Answer:

V2 = 35.967cm^3

Explanation:

Given data:

P1 = 0.2atm

P2 = 1.4atm

V1 = 250cm^3

V2 = ?

T1 = 10°C + 273 = 283K

T2 = 12°C + 273 = 285K

Apply combined law:

P1xV1/T1 = P2xV2/T2 ...eq1

Substituting values:

0.2 x 250/283 = 1.4 x V2/285

Solve for V2:

V2 = 14250/396.2

V2 = 35.967cm^3

7 0
2 years ago
How many moles are in 158 grams of LiCl
Usimov [2.4K]

Answer:

Number of moles = 3.73 mol

Explanation:

Given data:

Mass of LiCl = 158 g

Number of moles = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of LiCl = 42.4 g/mol

Number of moles =  158 g / 42.4 g/mol

Number of moles = 3.73 mol

7 0
3 years ago
9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M
antiseptic1488 [7]

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

5 0
3 years ago
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