Question

LE

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2. Valproic acid has a pKa of 4.8. Determine the pH of a 0.10 % w/v solution of valproic acid
(mwt 144.21).
(6 marks)
​

Answer 1

To determine the pH of a 0.10% w/v solution of valproic acid, we need to consider its pKa value. The pKa of valproic acid is 4.8. Since the pKa is higher than 2, valproic acid can be considered a weak acid. The pH of the solution can be calculated using the concentration of H+ ions in the solution.

Step-by-step explanation:

To determine the pH of a 0.10% w/v solution of valproic acid, we need to consider its pKa value. The pKa of valproic acid is 4.8. Since the pKa is higher than 2, valproic acid can be considered a weak acid. Weak acids partially dissociate in solution, releasing H+ ions.

To calculate the pH, we need to determine the concentration of H+ ions in the solution. We can use the equation pH = -log[H+]. To convert the percentage concentration to molarity, we can assume that the density of the solution is 1 g/mL. Therefore, 0.10% w/v is equal to 0.10 g/L. With a molar mass of 144.21 g/mol, we can calculate the molarity:

1. 0.10 g/L / 144.21 g/mol = 6.93 x 10^⁻⁴ M.
2. The concentration of H+ ions is equal to the concentration of valproic acid since it is a weak acid. Therefore, [H+] = 6.93 x 10^⁻³ M.
3. Finally, the pH can be calculated as:

pH = -log(6.93 x 10^⁻⁴) = 3.16.

Answer 2

pH=2.16

Step-by-step explanation:

Hello,

In this case, the first step is to compute the molarity of the 0.10%-w/v valproic acid solution by assuming 100 mL of solution:

`M=(0.10g)/(100mL)*(1mol)/(144.21g) *(1000mL)/(1L)\\ \\M=6.934x10^(-3)M`

Moreover, the equilibrium expression for the valproic acid (a weak one) is written as follows:

`Ka=([H^+][A^-])/([HA])`

Whereas HA represents the valproic acid and A⁻ its conjugate base. Thus, by computing Ka via its pKa and writing the aforementioned equilibrium expression in terms of
`x` we obtain:

`10^(-4.8)=1.585x10^(-5)=(x*x)/(6.934x10^(-3)-x)`

Thus, solving for
`x`, which also equals the concentration hydrogen ions, we obtain:

`x=[H^+]=6.93399x10^(-3)M`

Therefore, the pH is:

`pH=-log([H^+])=-log(6.93399x10^(-3))\\\\pH=2.16`

Best regards.