Question

12. A buffer that contains 100.0 mL of 0.150 M HIO (Ka-3.2 x 10-11) and 0.200 g of Na10 (165.89 g/mol) is prepared. (a) Calculate the pH of the buffer. (b) Calculate the pH after 2.00 ml. of 0.500 M NaOHl s added to 1000 mlL of the bufer prepared in Part (a).

Answer 1

Calculating pKa: Given K3.2x10 11 Ka -3.20E-11 pKa = -log [ Ka ] p' - - log [ 3.2x1011 ] 10.49 Calculating Concentration: Given, mass of NalO 0.200 molar mass of NalO 165.89 g/mol Volume of solution= Volume of solution= Concentration - 100 mL 0.1 mass molar mass x Volume in L 0.200 Concentration of NalO - 165.9 × 0.100 0.012 M Concentration of NalO-

weak acid HIO Conjugate base NalO HIO pKa-10.49 I NalO0.012 M I HIO 0.150 M Calculating pH: According to Henderson-Hasselbalch equation Conjugate base weak acid NalO HIO 1 pHpKa log pH -pKa log pH= 10.49 + log pH 9.40 10.0121 [ 0.150 1

Calculating initial number of moles: molariy of HIO 0.150 M volume of HIO = 100.0 mL 0.100 L number of moles of HIO? number of moles of HIOMolarity volume numberof moles of HIO = 0.150 x 0.100 numberof moles of HIO = 0.015 mol after 2 mL of NaOH added to 100.0 mL Calculating number of moles added: molariy of NaOH 0.500 M volume of NaOH= 2.00 mL = 0.002 L number of moles of NaOH? number of moles of NaOHMoaity volume numberof moles of NaOH = 0.500 × 0.002 number of moles of NaOH0.001 mol Number of moles of HIO0.015 0.0010.014 mol Number of moles of NalO0.0120.001 0.013 mol

Construct ICA table to obtain their equilibrium concentrations: Reaction NalO concentration ( mol ) 0.015 change ( mol ) -0.001 equilibrium ( mol)0.014 0.012 0.001 +0.001 0.013 Calculating pH: According to Henderson-Hasselbalch equation [ salt acid pHpKa + log pH-pKalog pH - 10.49 log pH = 10.465 NalO 1 [ HIO [ 0.013 1 [ 0.014 I