The pH = 2.41
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :
[tex]\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839[/tex]
Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
[tex]\tt [H^+]=\sqrt{Ka.M}[/tex]
Input the value :
[tex]\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41[/tex]
White vinegar is a 5.0% by mass solution of acetic acid in water and if the density of white vinegar is 1.007 g/cm³ then the pH of solution is 4.8.
pH of any solution will be calculated as:
pH = - log[H⁺], where
[H⁺] = concentration of H⁺ ions
In this question we will calculate the concentration of H⁺ ions as:
[H⁺] = √Ka.M, where
Ka = acid dissociation constant for acetic acid = 1.8 x 10⁻⁵ (known)
M is molarity and this can be calculated as:
M = (% mass × density × 10) / Molar mass of acid
Given that,
% mass = 5.0% by mass solution of acetic acid
density = 1.007 g/cm³
Molar mass of acetic acid = 60
Now putting all these values on the above equation, we get
M = (5 × 1.007 × 10) / 60 = 0.839
Now, we calculate the concentration of H⁺ ions by putting values on that equation as:
[H⁺] = √(1.8 x 10⁻⁵)(0.839)
[H⁺] = 1.5 x 10⁻⁵
So, pH of the solution will be calculated as:
pH = -log(1.5 x 10⁻⁵)
pH = -(-4.8)
pH = 4.8
Hence, pH is 4.8.
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https://brainly.com/question/10313314
