pH of a Buffered Solution
How to calculate ph of this solution?
A solution made by mixing 120.00 mL of 0.950 M HNO2 with 55.00 mL of 0.850 M NaOH
millimoles HNO2 = mL x M = 120.00 mL x 0.950 M = 114.
millimoles NaOH = 55.00 x 0.850 = 46.75
.......................HNO2 + NaOH ==> NaNO2 + H2O
I......................114..............0.................0............0
add.....................................46.75..................................
C...................-46.75.........-46.75.........46.75.....................
E...................67.25.................0..........46.75....................
total volume = 120.00 mL + 55.00 mL = 175.00 mL
(HNO2) = acid = a = 67.25 mmols/175 mL = ?
(NaNO2) = base = b = 46.75/175.00 = ?
Plug these into the HH equation and solve for pH. HH equation is
pH = pKa HNO2 + log (b/a)
Post your work if you get stuck.