1) A buffer system contains 0.24 M NH4+ and 0.12 M NH3. p K a of NH4+ is 9.25 Ho
ID: 1016949 • Letter: 1
Question
1) A buffer system contains 0.24 M NH4+ and 0.12 M NH3. pKa of NH4+ is 9.25
How many moles of NaOH must be added to 1.00 L of this solution to increase the pH to 9.2?
2) Titration of a 19.0 mL solution of KOH requires 17.0 mL of 0.0210 M H2SO4 solution
What is the molarity of the KOH solution?
3) A 0.17 M solution of HCl is used to titrate 28.0 mL of a Ca(OH)2 solution of unknown concentration
If 170 mL of HCl is required, what is the concentration (in molarity) of the Ca(OH)2 solution?
Please show your work and how you get the answer so I can follow along, thanks!
Explanation / Answer
(1) Number of moles of NH3 = Molarity x volume in L = 0.12 M x 1.00L = 0.12 mol
Number of moles of NH4+ = Molarity x volume in L = 0.24 Mx1.00L = 0.24 mol
NH4+ + OH- ---> NH3 + H2O
Initial 0.24 0 0.12
add 0 +c 0
Change -c -c +c
Equib 0.24-c 0 0.12+c
According to Henderson pH = pKa + log([salt]/[acid])
9.2 = 9.25 + log((0.12+c)/(0.24-c))
Solving for c we get c = 0.047 moles
So the number of moles of NaOH must be added is 0.047 mol
(2) Number of moles of H2SO4 is , n = Molarity x volume in L
= 0.0210 M x 0.017 L
= 3.57x10-4 mol
The balanced reaction is 2KOH + H2SO4 ----> K2SO4 + 2H2O
From the balanced reaction ,
1 moles of H2SO4 reacts with 2 moles of KOH
3.57x10-4 moles of H2SO4 reacts with M moles of KOH
M = ( 3.57x10-4 x2) /1
= 7.14x10-4 mol
So Molarity of KOH , M = Number of moles / volume in L
= 7.14x10-4 mol / 0.019 L
= 0.0376 M
Therefore the molarity of KOH is 0.0376 M
(3) Number of moles of HCl , n =Molarity x volume in L
= 0.17M x 0.170 L
= 0.0289 moles
The balanced reaction is : 2HCl + Ca(OH)2 ----> CaCl2 + 2H2O
From the balanced reaction ,
2 moles of HCl reacts with 1 mole of Ca(OH)2
0.0289 moles of HCl reacts with N moles of Ca(OH)2
N = ( 0.0289x1) / 2
= 0.01445 moles
So Molarity of Ca(OH)2 is , M = Number of moles / volume in L
= 0.01445 mol / 0.028 L
= 0.516 M
Therefore the required Molarity is 0.516 M
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