homework

[sarahtang4B]

On 7th edition problem 6B11, I'm confused how to do it, for a)ii) why do you multiply by 55ml/5ml?

[Chem_Mod]

To solve you will want to set up and balance the reaction Na₂O + H₂O 2NaOH. Then this solution of NaOH is diluted by taking 5 mL or 0.005 L of the solution and diluting it to a final volume of 0.5 L. Using the given pH of this final solution to be 13.25 we can calculate the pOH by taking 14-13.25=0.75. Then to find the concentration of OH^- we take 10^-0.75=0.18 M. Since NaOH is a strong base, the concentration of OH^- in solution will be equal to the concentration of NaOH so the answer to i) is 0.18M. For part ii) we would have to work backwards from this concentration as the dilute and use M₁V₁=M₂V₂ and when we substitute in we get (0.005L)M₁=(0.5L)(0.18M) this gets M=17.8M which is the concentration of NaOH in the original solution. For part b) it wants the mass of Na₂O, so we first multiply the concentration by the volume to get the moles of NaOH 17.8 M * 0.2 L =3.556 moles. Then using the balanced equation we find that for every 2 mol of NaOH there was 1 mol of Na₂O so divide 3.556 by 2 to get that there are 1.778 mol of Na₂O which when using the molar mass of 62 g/mol we get that there was 110 g added.