What is the pH of 80.0 mL of a solution which is 0.21 M in IO- and 0.41 M in HIO

Question

What is the pH of 80.0 mL of a solution which is 0.21 M in IO- and 0.41 M in HIO? For HIO use Ka =2.3 times 10-11. pH =

Explanation / Answer

when the Ka = 2.3 × 10–11 then the pKa = 10.638 henderson hasselbalch: http://en.wikipedia.org/wiki/Henderson-H… pH = pKa + log[ (IO-) / (HIO) ] pH = 10.638 + log[ (0.27) / (0.43) ] pH = 10.638 + log[ (0.6279) ] pH = 10.638 + (-0.202) pH = 10.4359 sorry it is still, by any method to 2 sig figs... pH = 10.44 ========================== 0.1000 L of 0.100 M in NH3(aq) = 0.0100moles NH3 and 0.1000 L of 0.100 M in NH4Cl(aq). = 0.0100moles NH4)+1 has initially: NH3 (aq) NH4+ & OH- 0.0100 mol 0.0100 mol & X mol the Ka = [NH4] [OH-] / [NH3(aq)] 1.8 X 10^-5 = [0.100 M] [OH-] / [0.100M] [OH-] = 1.8 X 10^-5 pOH = 4.745 pH = 9.255 my Kb is only good for 2 sig figs, unless your Kb was good for more sig figs the initial pH should be 9.26 ============================== when 0.00400 L of 0.100 M HCl(aq) is added = 0.000400 moles of H+ are added the equilib shifts to the left => NH3 (aq) => => NH4+ & OH- 0.0100 - 0.0004 => 0.0100 + 0.0004mol & X we now have: NH3 (aq) => => NH4+ & OH- 0.0096 mol / 104 ml => 0.0104mol/ 104 ml & [OH-] the Ka = [NH4] [OH-] / [NH3(aq)] 1.8 X 10^-5 = [0.0104mol/ 0.104 L] [OH-] / [0.0096mol/ 0.104L] [OH-] = (1.8 X 10^-5 ) (0.923) [OH-] = 1.66 X 10^-5 pOH = 4.779 pH = 9.221 ============================= Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added 9.221 - 9.255 = it drops by 0.034 rounded to 2 sig figs earlier on, or now gives delta pH = - 0.03 ======================== Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added = 0.000400 moles of OH- are added the equilib shifts to the right NH4+ & OH- 0.0104 mol / 104 ml => 0.0096mol/ 104 ml & [OH-] the Ka = [NH4] [OH-] / [NH3(aq)] 1.8 X 10^-5 = [0.0096mol/ 0.104 L] [OH-] / [0.0104mol/ 0.104L] [OH-] = 1.95 X 10^-5 pOH = 4.710 pH = 9.290 d pH = 9.290 - 9.255 = 0.03 it is a difficult call when something is so close to be rounded off to either 0.03 or 0.04, depending upon how many digits someone decided to carry through the problem

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