Question

13.5 g of an acid HA of molecular mass 135 was dissolves in 10 liters of aqueous solution . calculate the ph of the solution, assuming the acid to be completely dissociated

Answer 1

Morality = (13.5/135)/10 = 0.01 M
It seems to be monoprotoic acid.
HA === H+ + A-

So [ H+] = 0.01M.
pH = - log [H+] = 2

Answer 2

Answer: 2

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{\text{given mass of compound}}{\text{molar mass of compound}\times \text{volume of solution is L}}$

$Molarity=\frac{13.5g}{135g/mol\times 10L}=0.01M$

$HA\rightarrow H^++A^-$

if acid dissociates completely ,1 mole of HA gives 1 mole of $H^+$

0.01 mole of HA gives 0.01 mole of $H^+$

$pH=-log[H^+]$

$pH=-log[0.01]=2$