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Question

A student measured out a 15.0 mL aliquot (portion) of juice into an Erlenmeyer flask. She added 3 drops of Phenolphtalein indicator to the solution. It took 27.30 mL of 0.1003 M NaOH to reach the endpoint of the titration. 

(a) Determine the moles of NaOH required for the titration. Use sig figs 

(b) What is the net ionic equation for an acid-base neutralization reaction?

(c) Determine the total of moles of H^+ in the juice aliquot. Use sig figs. 

Answer 1

Here is how to solve this:

(a). To find the moles of NaOH, we use dimensional analysis as follows:

27.30 ml NaOH x 1 L / 1000 ml = 0.02730 L of NaOH

0.02730 L NaOH x 0.1003 mols NaOH / L = 0.002738 moles of NaOH required (4 sig.figs.)

(b). To write the net ionic equation, we will first write the complete molecular equation, and then remove the spectator ions. Unfortunately, the question does NOT state the identity of the acid, so we will have to make an assumption. We will assume that the juice in question contains a mono-protic acid, meaning that this acid has only 1 ionizable hydrogen. We will refer to it as HA. We will also assume that it is a strong acid (not a weak acid), and so it ionizes 100%.

HA(aq) + NaOH(aq) ==> H2O(I) + NaA(aq) ... complete molecular equation

H+(aq) + A-(aq) + Na+(aq) + OH-(aq) ==> H2O(l) + Na+(aq) + A-(aq) ... total ionic equation

H+(aq) + OH-(aq) ==> H2O(l) ... NET IONIC EQUATION

(c). To determine moles of H+ in the juice, we will use the balanced equation (net ionic) above, and the moles of NaOH determined in part (a) above, as follows:

H+(aq) + OH-(aq) ==> H2O(l)

moles of OH- used = 0.002738 moles of NaOH

moles of H+ present = 0.002738 moles of NaOH x 1 mole H+ / 1 mol NaOH = 0.002738 moles of H+ (4 s.f.)