A) For 480.0 mL of pure water, calculate the initial pH and the final pH after a

Question

A) For 480.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of HCl. Express your answers using two decimal places separated by a comma.

B) For 480.0 mL of a buffer solution that is 0.150 M in HC2H3O2 and 0.135 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.010 mol of HCl.

Express your answers using two decimal places separated by a comma.

C) For 480.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.170 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of HCl.

Express your answers using two decimal places separated by a comma.

Explanation / Answer

A) pH of pure water is 7, meaning the [H+] = [OH-], by adding 0.01 moles of HCl to the 0.48L water, you create a 0.01moles / 0.48L = 0.021M solution of HCl, which gives you a [H+] of 0.021M
pH = -log[H+] = -log[0.021M] = 1.67

C)

Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37
pOH = 3.37 + log 0.17 / 0.185 = 3.33
pH = 14 - 3.33 =10.67 ( initial pH)
moles CH3CH2NH3+ = 0.17 x 0.480 L=0.0816
moles CH3CH2NH2 = 0.185 x 0.480 L=0.0888
CH3CH2NH2 + H+ = CH3CH2NH3+
moles CH3CH2NH3+ = 0.0816 + 0.010 =0.0916
moles CH3CH2NH2 = 0.0888 - 0.010 =0.0788
concentration CH3CH2NH3+ = 0.0916 / 0.480 =0.191 M
concentration CH3CH2NH2 = 0.0788 / 0.480=0.164 M
pOH = 3.37 + log 0.191 / 0.164=3.43
pH = 10.57

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