Solution:
Benzoic acid is C6H5COOH
In finding pH
C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)
Ka = 6.45 x 10^-6
[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.
Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6
Now,
According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x
x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0
enter a = 1, b = 0.00000645, c = 0.0000323
x = 5.68 x 10^-3 = 0.00568 M expression is [C6H5COOH] = 0.5 M is the correct answer.
[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25
This is the required solution.
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Answer:
C₆H₆
Explanation:
We need to find the molecular formula of a compound of carbon (C) and hydrogen (H), so what <em>we need to find out is the number of atoms of C and of H in the molecule.</em> We know:
- molar mass = 78.1 g/mol
- C% = 92.3% = 92.3 g C / 100 g compound
So, in 1 mol of compound, 92.3% of the mass corresponds to Carbon:
<u>mass of C / mol of compound</u> = molar mass × C% = 78.1 g/mol × 92.3/100 = <u>72.1 g/mol</u>
<u>moles of C</u> = mass C / molar mass C = 72.1 g / 12.011 g/mol
moles of C = 6 moles of C per mol of compound
If 72.1 g in a mol of compound are Carbon atoms, the difference between the molar mass and the mass of Carbon atoms will correspond to H atoms in 1 mol of compound:
<u>mass of H / mol of compound</u> = molar mass - mass of C/mol
mass of H = 78.1 g / mol - 72.1 g /mol = <u>6.0 g/mol of compound</u>
<u>moles of H</u> = mass H / molar mass H = 6.0 g / 1.008 g/mol
moles of H = 6.0 moles of H per mol of compound
<em>So</em><em> one mol of compound has 6 moles of C and 6 moles of H.</em>
The molecular formula is then written as C₆H₆
