0.0155L of a 0.505M solution of NaOH is combined with 0.0482L of a 0.0175M solution of HCl. What is the pH of the combined solution?
please show all steps!!
first let us calculate moles of each entity
NaOH moles = concentration * volume = .505 * .0155
= 0.0078275 moles
HCl moles = concentration * volume = .0175 * .0482 = 0.0008435 moles
we can find that moles of NaOH > moles of HCl
remaining moles of NaOH = 0.0078275 - 0.0008435 = 0.006984 moles
conc of NaOH = moles / volume = 0.006984/ (.0155+.0482) = 0.1096 M
Since NaOH is a strong acid we assume that it dissociates completely
So conc of OH- = Cnonc of NaOH = 0.1096
pOH= -log(0.1096) = .96
pH = 14 - pOH = 13.04 is the pH
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