The pH of a solution obtained by mixing 100cm\(^3\) of a 0.1M HCl solution with 100cm\(^3\) of a 0.2M solution of NaOH is?
Calculate Moles of Reactants:
Moles of HCl = Molarity x Volume in dm\(^3\) = 0.1 M x 0.1 dm\(^3\) = 0.01 moles
Moles of NaOH = Molarity x Volume in dm\(^3\) = 0.2 M x 0.1dm\(^3\) = 0.02 moles
Determine the Excess Reagent:
The neutralization reaction is a 1:1 ratio: HCl + NaOH → NaCl + H\(_2\)O
Excess NaOH = 0.02 - 0.01 = 0.01 mole
Calculate the Hydroxide Ion Concentration[OH\(^-\)]
Total Volume of the mixture = 100 cm\(^3\) + 100cm\(^3\) = 200 cm\(^3\) = 0.2dm\(^3\)
[OH\(^-\)] = \(\frac{0.01}{0.2}\) = 0.05M
Calculate pOH and pH
pOH = - Log[OH\(^-\)] = - Log[0.05]
= 1.3
But pOH + pH = 14
1.3 + pH = 14
pH = 14 - 1.3
= 12.7
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