to make 500.0 mL of solution? s the pH of solution when 25.5 g NaC2H,O2 is added
ID: 1026669 • Letter: T
Question
to make 500.0 mL of solution? s the pH of solution when 25.5 g NaC2H,O2 is added to 0.550 M HC HsO e pH of after adding 0.0060 mol HCI to 0.300 L of a buffer solution that is 0.250 M HC2H O2 and 0.560 M NaC2H3O2? (k) What is the pH of after adding 0.0060 mol NaOH to 0.300 L of a buffer solution that is 0.250 M HC2Hs02 and 0.560 M Na C2H302? (C) What is the pH of a solution containing 50.00 mL of 0.100 M HCI to which 30.00 mL of 0.100 M NaOH has been added? (m) What is the pH of a solution containing 50.00 mL of 0.100 M HCl to which 50.00 (n) What is the pH of a solution containing 50.00 mL of 0.100 M HCl to which 60.00 o) What is the pH of a solution that contains 50.00 mL of 0.10 M NH3 to which ) What is the pH of a solution that contains 50.00 mL of 0.10 M NHs to which What is the pH of a solution that contains 50.00 mL of 0.10 M NHs to which mL of 0.100 M NaOH has been added? mL of 0.100 M NaOH has been added? 30.00 mL of 0.10 M HCI has been added? 50.00 mL of 0.10 M HCI has been added? 60.00 mL of 0.10 M HCI has been added?Explanation / Answer
1.
mass of NaC2H3O2 = 25.5 grams
molar mass of NaC2H3O2= 82.0 gram/mole
nubber of moles of NaC2H3O2= 25.5/82.0 = 0.311 moles
HC2H3O2 = 0.550M of 500 ml
number of moles fo HC2H3O2 = 0.550Mx0.500L= 0.275 moles
PKa of HC2H3O2= 4.76
PH= PKa + log[salt]/[acid]
PH= 4.76 + log(0.311/0.275)
PH=4.81
J) number of moles of HCl= 0.0060 moles
number of moles of NaC2H3O2 = 0.560M x0.3L = 0.168 moles
number of moles of HC2H3O2 = 0.250Mx0.3L= 0.075 moles
after addition o f HCl
number of moles of HC2H3O2 = 0.075 + 0.0060 =0.081 moles
number of moles of NaC2H3O2 = 0.168 - 0.0060 = 0.162 moles
PH= 4.76 + log(0.162/0.081)
PH= 5.06
K)
number of moles of NaOH= 0.0060 moles
number o fmoles of HC2H3O2 = 0.250Mx0.300L= 0.075 moles
number of moles of NaC2H3O2 = 0.560M x0.3L = 0.168 moles
after additon of NaOH
number of moles of HC2H3O2 = 0.075 - 0.006 =0.069 moles
number of moles of NaC2H3O2 = 0.168 + 0.006= 0.174 moles
PH= 4.76 + log(0.174/0.069)
PH= 5.16
l)
number of moles of HCl=0.1Mx0.050L=0.005moles
number of moles of NaOH= 0.1Mx0.03L=0.003 moles
number of moles of HCl is greaterthan the NaOH.
so the solution has acidic nature ;
remaining number of moles of HCl = 0.005 - 0.003 = 0.002 moles
total volume = 50.0+30.0= 80.0ml=0.080L
[H+]= 0.002/0.080=0.025 M
-log[H+]= -log(0.025)
PH= 1.60
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