Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL

Question

          Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Delta pH=

          Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09

          Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Delta pH=

          Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09

     Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Delta pH=

          Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09

Calculate the change in pH when 6.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Delta pH=

          Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09

          Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09
     Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09
Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Delta pH=

answers are NOT .04, .09

Explanation / Answer

pOh = pkb + log [NH4Cl]/[NH3]

pOH = 4.75 + log ( 0.1/0.1) = 4.75 , pH = 9.25

when HCl of 0.1x0.006 added then [NH3] = ( 0.1x0.1 -0.0006)/0.106 = 0.0094/0.106

[NH4Cl] = ( 0.x0.1+0.0006)/0.106 = 0.0106/0.106

pOH = 4.75 + log ( 0.0106/0.0094) = 4.8 , pH = 14-4.8 = 9.2

pH change = 9.25-9.2 = 0.05

when 0.1 x0.006 moles NaOH added

[NH3] = ( 0.1x0.1+0.0006)/0.106 = 0.0106/0.106

[NH4Cl] = ( 0.01-0.0006)/0.106 = 0.0094/0.106

pOH = 4.75 + log ( 0.0094/0.0106) = 4.7 , pH = 14-4.7 = 9.3

pH change = 9.25 -9.3 = -0.05

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