Calculate the ph after addition of 0.00,5.00,15.00,25.00,40.00,45.00,49.00,50.00

Question

Calculate the ph after addition of 0.00,5.00,15.00,25.00,40.00,45.00,49.00,50.00,51.00,55.00,and 60.00ml of 0.1000m NaOH in the titration of 50ml of A. 0.1000M HNO2 B. 0.1000M pyridium chloride C. 0.1000M acetic acid Calculate the ph after addition of 0.00,5.00,15.00,25.00,40.00,45.00,49.00,50.00,51.00,55.00,and 60.00ml of 0.1000m NaOH in the titration of 50ml of A. 0.1000M HNO2 B. 0.1000M pyridium chloride C. 0.1000M acetic acid A. 0.1000M HNO2 B. 0.1000M pyridium chloride C. 0.1000M acetic acid

Explanation / Answer

A: For HNO2, pKa = 3.398

pH When 0.00 mL of HNO2 is added: HNO2 is a weak acid and the pH of a weak acid can be calculated as

The chemcial equation for the dissociation of HNO2 is

------------- HNO2 ------- > NO2-(aq) + H+(aq) ; Ka = 4.0*10-4

init.con: 0.100M ----------- 0 M -------- 0 M

change: - 0.100x M, +0.100xM, +0.100xM

eqm.con:0.100(1 - x)M, 0.100xM, 0.100xM

Ka = 4.0*10-4 = [NO2-(aq)]*[H+(aq)] / [HNO2] = (0.100xM * 0.100xM) / 0.100(1 - x)M

since x <<1, (1 - x) is nearly equals to 1.

=> x = underroot( 4.0*10-3) = 0.06325

  [H+(aq)] =  0.100x M = 0.100 x 0.06325 = 0.006325

=> pH = -log[H+(aq)] = - log( 0.006325) = 2.20 (answer)

pH when 5 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.005 L = 0.00050 mol

0.00050 mol NaOH will react with 0.00050 mol HNO2 to form 0.00050 mol NaNO2.

Hence moles of NaNO2 = 0.00050 mol

moles of HNO2 = 0.00450 mol

HNO2 and NaNO2 now acts as buffer solution whse pH can be calculated from Hendersen equation

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.00050 mol / 0.00450 mol) = 2.44 (answer)

pH when 15 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.015 L = 0.0015 mol

Hence

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.0015 mol / 0.00350 mol) = 3.03 (answer)

pH when 25 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.025 L = 0.0025 mol

Hence

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.0025 mol / 0.00250 mol) = 3.398 (answer)

pH when 40 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.040 L = 0.0040 mol

Hence

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.0040 mol / 0.0010 mol) = 4.00 (answer)

pH when 45 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.045 L = 0.0045 mol

Hence

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.0045 mol / 0.0005 mol) = 4.352 (answer)

pH when 49 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.049 L = 0.0049 mol

Hence

pH = pKa + log[NaNO2] / [HNO2]

=> pH = 3.398 + log(moles of NaNO2 / moles of HNO2] [ Since vol. remains same for both]

=> pH = 3.398 + log(0.0049 mol / 0.0001 mol) = 5.088 (answer)

pH when 50 mL of 0.1000 M NaOH is added:

initial moles of HNO2 = MxV = 0.1000 mol/Lx 0.050 L = 0.0050 mol

moles of NaOH added = MxV = 0.1000 mol/Lx 0.049 L = 0.0050 mol

Hence equivalence point is achieved at 50 mL NaOH

naWe can calculate the pH at equivalence point by salt hydrolysis

[NaNO2] = moles of NaNO2 / total volume = 0.0050 mol / 0.100 L = 0.05 M

Hence pH = (1/2)*[pKw+ pKa + log[HNO2]]

=> pH = (1/2)*[14 + 3.398 + log(0.05)] = 8.05 (answer)

pH when 51 mL of 0.1000 M NaOH is added:

Moles of excess NaOH = 0.100 mol/L x 0.001 L = 0.0001 mol

total volume = 50 mL + 51 mL = 101 mL = 0.101 L

=> [OH-] = 0.0001 mol / 0.101 L = 9.901x10-4 M

=> pOH = - log[OH-] = 3.004

=> pH = 14 - 3.004 = 10.996 (answer)

Similarly we can calculate the pH in a same manner for the rest

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