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For 500.0 mL of a buffer solution that is 0.185 molL−1 in C2H5NH2 and 0.175 molL−1...

For 500.0 mL of a buffer solution that is 0.185 molL−1 in C2H5NH2 and 0.175 molL−1 in C2H5NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

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Answer #1

pKb = 3.25

now use the formula

pOH = pKb + log(salt/acid)
pOH = 3.25 + log (0.175 / 0.185)

pOH = 3.25-0.024

pOH = 3.23

now pH = 14-pOH

pH = 14-3.23 = 10.77 initial pH

no of moles of C2H5NH2 = 0.185 mol/L x 0.5L = 0.0925 moles

no o fmoles of C2H5NH3Cl = 0.175 mol/L x 0.5L = 0.0875 moles

now you added 0.02 moles of HCl so this will react with 0.02 moles of C2H5NH2 and will give the 0.02 moles of C2H5NH3Cl

after addition from C2H5NH2 0.02 moles will decrease = 0.0925 - 0.02 = 0.0725 moles

after addition 0.02 moles of HCl addition C2H5NH3Cl will increase 0.0875 + 0.02 = 0.1075 moles

concentration of C2H5NH2 = 0.0725 moles / 0.5 L = 0.145 mol/L

concentration of C2H5NH3Cl = 0.1075 / 0.5 = 0.215 mol/L

now

pOH = 3.25 + log(0.215/0.145)

pOH = 3.25 - 0.171

pOH = 3.08

pH = 14-pOH

pH = 14-3.08

pH = 10.92 final pH

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