Question

Please help. I am studying for an upcoming exam and I am having a hard time doing these two questions.

21-90. What is the pH of the solution that results from the addition of 25.0 mL of 0.200 M KOH(aq) to 50.0 mL of 0.150 M HNO,(aq)?

Answer 1

2I-90.

molarity (M) = moles/L

KOH + HNO2 = KNO2 + H2O

1 mole of KOH reacts with 1 mole of HNO2.

moles of KOH = volume in L x molarity = 0.025 L x 0.200 moles/L = 0.005 moles

moles of HNO2 = 0.050 L x 0.150 moles/L = 0.0075 moles

so moles of HNO2 remaining after neutralisation = 0.0075 - 0.005 = 0.0025 moles

total volume of solution = 0.025 L + 0.050 L = 0.075 L

concentration of HNO2 = 0.0025 moles/0.075 L = 0.0333 M

Ka for HNO2 = 4.6 x 10^-4

HNO2 = H+ + NO2-

0.0333 - x x x

Ka = x2/0.0333-x

0.0333 - x = 0.0333 (because x is very small )

4.6 x 10^-4 = x2/0.0333

x = 3.9138 x 10^-3 M = [H+]

pH = -log[H+] = -log[3.9138 x 10^-3 M] = 2.41

2I-67.

oxalic acid = H2C2O4

H2C2O4 + NaOH = H2O + NaHC2O4 till first equivalence point

NaHC2O4 + NaOH = H2O + Na2C2O4

moles of NaOH = .02777 L x 1.00 moles/L = 0.02777 moles

so till first equivalence point moles of NaOH and moles of oxalic acid are equal

mole of oxalic acid = 0.02777 moles

moles = mass of oxalic acid/ molecular mass of oxalic acid

0.02777 moles = 2.500 g/molecular mass of oxalic acid

molecular mass of oxalic acid = 2.500 g/0.02777mole = 90.0252 g/mole

...........................

if you have any question please feel free to put down in comment box...

if you have found it useful please give it a thumbs up