Question

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!!

Answer 1

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