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Butoxors [25]
2 years ago
7

The ph of a 0.175 m aqueous solution of a weak acid is 3.52. what is ka for this acid?

Chemistry
1 answer:
Naddika [18.5K]2 years ago
3 0

Hey there!

pH = - log [ H⁺  ] = 3.52

[ H⁺ ] = 10^-pH

[ H⁺] = 10^ ( -3.52 )

[H⁺] = 3.02*10⁻⁴ M

[HA] =0.175 M

Therefore:

Ka = [ H⁺]* [A⁻] / [ HA]

Ka =  (3.02*10⁻⁴)² /  0.175

Ka = 9.1204*10⁻⁸ / 0.175

Ka = 5.2*10⁻⁷


Hope that helps!


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A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.
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Answer:  The empirical formula and the molecular formula of the organic compound is CHO and C_4H_4O_4 respectively.

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The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

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Mass of H_2O= 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, =\frac{12}{44}\times 21.71=5.921g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, =\frac{2}{18}\times 5.926=0.658g of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53  g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles

Mass of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.493}{0.493}=1

For H =\frac{0.658}{0.493}=1

For O=\frac{0.658}{0.493}=1

The ratio of C : H: O =  1: 1: 1

Hence the empirical formula is CHO.

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4

Thus molecular formula = n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4

6 0
2 years ago
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