A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50.0 mL of 0
ID: 794440 • Letter: A
Question
A buffer solution is prepared by mixing 50.0 mL of 0.300 M NH3 with 50.0 mL of 0.300 M NH4Cl. the pKb is 4.74.
a) Calculate the [NH3] and [H4Cl] in the buffer solution. Calculate the pH of the buffer solution.
b) 7.50 mL of 0.125 M HCl is added to the 100.0 mL of the buffer solution. Calculate the new [NH3] and [NH4Cl] for the buffer solution. Calculate the new pH of the solution
c) 7.50 mL of 0.125 M NaOH is added to the 100.0 mL of the buffer solution. Calculate the new [NH3] and [NH4Cl] for the buffer solution. Calculate the new pH of the solution.
***Please give a thorough step-by-step solution for each part. Please include all units in your work so I can keep track of what everything is measured in! Thanks!***
Explanation / Answer
pOH=pKb+log[salt]/[base]
equation can be written in terms of no. of moles
pOH = pKb + log[no.of moles of salt/ no.of moles of base]
a)
[NH3]=50*0.3=15
[NH4Cl]=50*0.3=15
pH=14-pOh=14-4.74=9.26
b) [HCl]=7.5*0.125=0.9
by the addition of 0.9 moles of HCl, the same amount of NH3 will be neutralised.
therefore,
[NH3]=14.1
[NH4Cl]=15.9
pH=14-4.74-log(15.9/14.1)=9.23
c) [NaOH] = 7.5*0.125=0.9
by the addition of 0.9 moles of NaOH, the same amount of OH- concentration will be increased.
therefore,
[NH3] = 15.9
[NH4Cl] = 14.1
pH=14-4.74-log(14.1/15.9) = 9.31
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