2. Calculate the pH of the resulting solutions: a) 35.0 ml of 0.300 M HCl added

Question

2. Calculate the pH of the resulting solutions:
a) 35.0 ml of 0.300 M HCl added to 20.0 ml 0.500M NaOH
b) 30.0 ml of 0.545 M KOH added to 30.0 ml of 0.545 M HC2H3O2
c)55.0 ml of 0.880 M HCl added to 40.0 ml of 1.21 M NaC2H3O2

Please explain the steps so I can understand.

Explanation / Answer

part a). HCl + NaOH -----> NaCl + H2O If equal amounts and molarities of NaOH and HCl were added they would make water which is neutal. This is not the case so we have to find out how much of one will be left over. C (HCl) = n/v 0.300 = n(HCl) /0.035 n(HCl) = 0.0105 C (NaOH) = n/v 0.500 = n/0.02 n(NaOH) = 0.01 mol As you can see there is more HCl than NaOH. n(HCl) left over = 0.0105-0.01 = 0.0005 mol We can now calculate the pH n(HCl) = c/v 0.0005 = c/0.055

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