Consider a 1.0-L solution that is 0.830 M HNO2 and 0.37 M NaNO2 at 25 °C. What is the pH of this solution before and after 0.011 moles of LiOH have been added? Ka of HNO2 is 4.5×10-4.
initial pH: 3.7, final pH: 2.98.
initial pH: 3.35, final pH: 2.98.
initial pH: 3.0, final pH: 3.68.
initial pH: 3.7, final pH: 3.72.
initial pH: 3.0, final pH: 3.01.
1)
Lets calculate the initial pH
Ka = 4.5*10^-4
pKa = - log (Ka)
= - log(4.5*10^-4)
= 3.347
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.347+ log {0.37/0.83}
= 3.00
2)
Lets calculate the final pH
mol of LiOH added = 0.011 mol
HNO2 will react with OH- to form NO2-
Before Reaction:
mol of NO2- = 0.37 M *1.0 L
mol of NO2- = 0.37 mol
mol of HNO2 = 0.83 M *1.0 L
mol of HNO2 = 0.83 mol
after reaction,
mol of NO2- = mol present initially + mol added
mol of NO2- = (0.37 + 0.011) mol
mol of NO2- = 0.381 mol
mol of HNO2 = mol present initially - mol added
mol of HNO2 = (0.83 - 0.011) mol
mol of HNO2 = 0.819 mol
Ka = 4.5*10^-4
pKa = - log (Ka)
= - log(4.5*10^-4)
= 3.347
since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.347+ log {0.381/0.819}
= 3.01
Answer:
initial pH: 3.0, final pH: 3.01.
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