- #1
terryds
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Homework Statement
500 mL KOH solution (0.1 M) is mixed with 500 mL HNO2 (0.1 M) so that the reaction becomes:
##KOH + HNO_2 \rightarrow KNO_2+H_2O##
If Ka of HNO2 is ##5 \times 10^{-4}##, determine the pH of the solution
2. The attempt at a solution
##KOH + HNO_2 \rightarrow KNO_2+H_2O##
Both KOH and HNO2 are in the same number of moles, which is 0.5 * 0.1 = 0.05 mol
So, the product (KNO2) is also 0.05 mol due to the same coefficient
And, the ionization reaction of KNO2 is
##KNO_2 \rightarrow K^+ + NO_2^-##
NO2 is also 0.05 mol due to same coefficient.
But, how to determine the pH??
I don't see how to get H+ or OH- into the reaction so I'll be able to calculate the pH...
I don't understand what the Ka of HNO2 matters since KOH and HNO2 has fully react becoming KNO2 and H2O (In other words, there are no more KOH and HNO2)