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HA ⇄ H⁺ + A⁻
so:
![\frac{[H^+][A^-]}{[HA]} = 1.5 x 10^{-5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%3D%201.5%20x%2010%5E%7B-5%7D%20%20)
and now:
= 1.5 x 10⁻⁵
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
so:
and now:
x is considered very small compared to 0.15
x² = 2.25 x 10⁻⁶
x = 1.5 x 10⁻³
So [H⁺] = 1.5 x 10⁻³
pH = - log [H⁺] = - log (1.5 x 10⁻³) = 2.83
<u>Answer:</u> The number of moles of weak acid is moles.
<u>Explanation:</u>
To calculate the moles of KOH, we use the equation:
We are given:
Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)
Molarity of the solution = 0.0969 moles/ L
Putting values in above equation, we get:
The chemical reaction of weak monoprotic acid and KOH follows the equation:
By Stoichiometry of the reaction:
1 mole of KOH reacts with 1 mole of weak monoprotic acid.
So, of KOH will react with =
of weak monoprotic acid.
Hence, the number of moles of weak acid is moles.