Asked by Anonymous
500 ml of a buffer solution with ph=2.10 must be prepared using .4 M HNO2 and solid KNO2. The ka value of HNO2 is 4e-3.
a.) What mass of KNO2 should be added to 3 L of the HNO2 to make the buffer?
b.) What is the buffer's pH after 150 ml of .5 M HNO3 is added?
a.) What mass of KNO2 should be added to 3 L of the HNO2 to make the buffer?
b.) What is the buffer's pH after 150 ml of .5 M HNO3 is added?
Answers
Answered by
DrBob222
Approx means I've estimated. You should go through yourself and get good answers.
pKa = approx 2.4
mols HNO2 = 0.5L x 0.4M = 0.2 = acid
pH = pKa + log (base)/(acid)
2.10 = 2.40 + log (x/0.2)
Solve for x = mols KNO2 = about 0.1 mol = ? grams.
b.
millimols HNO2 = 500 mL x 0.4M = 200
mmols HNO3 added = 150 x 0.5M = 75
mmols KBNO2 = 100 from above.
............NO2^- + H^+ ==> HNO2
I.......... 100......0.......200
added........0......75........0
C...........-75.....-75.......+75
E............25......0........275
Plug that E line (after you've recalculated everything to get the good numbers to use) into HH equation and solve for pH.
pKa = approx 2.4
mols HNO2 = 0.5L x 0.4M = 0.2 = acid
pH = pKa + log (base)/(acid)
2.10 = 2.40 + log (x/0.2)
Solve for x = mols KNO2 = about 0.1 mol = ? grams.
b.
millimols HNO2 = 500 mL x 0.4M = 200
mmols HNO3 added = 150 x 0.5M = 75
mmols KBNO2 = 100 from above.
............NO2^- + H^+ ==> HNO2
I.......... 100......0.......200
added........0......75........0
C...........-75.....-75.......+75
E............25......0........275
Plug that E line (after you've recalculated everything to get the good numbers to use) into HH equation and solve for pH.
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