Question

Question 1 20 points Save Answer You add 19.00 mL of 0.700 M NaOH to 50.00 mL of pure water, and to this mixture you then add 9.00 mL of 0.400 M HCL. What will be the pH of the resulting solution?

Answer 1

To calculate the pH of the solution we need to find the [H+],concentration of H+ or [OH-],concentration of OH-.

Millimoles = Molarity*Volume(in mL)

Millimoles of NaOH = Molarity of NaOH*Volume(in mL) = 0.7*19 = 13.3 millimoles

MIlimoles of HCl = Molarity of HCl*Volume(in mL) = 0.4*9 = 3.6 millimoles

Since HCl is an acid and NaOH is base, neutralization reaction will occur and form NaCl and water, HCl + NaOH → NaCl + H2O. As we can 1 mole of HCl will 1 mole of NaOH to form 1 mole of NaCl and 1 mole of water. In the solution there are 3.6 millimoles of HCl that will react with 3.6 milimoles of NaOH to take part in neutralization reaction. So there will left only NaOH in the solution as there will no more HCl to react with.

Remaining NaOH = 13.3 - 3.6 = 9.7 millimoles = 9.7*10-3 moles

Moles of OH- = Moles of NaOH = 9.7*10-3 moles

Final Volume of Solution = 50+19+9 =78mL

Molarity of OH-,[OH-] = Moles of OH- / Final Volume(in litres) = 9.7*10-3/ 78*10-3 = 0.1243M

pOH = -log10[OH-] = -log10[0.1243] = 0.9053

At room temperature(25 degree Celsius), pOH + pH = 14

Therefore pH = 14 - pOH = 14 - 0.9053 = 13.094

pH of the resulting solution is 13.094.