Titration and M = n/V

Question

If the lemon juice used in the lab has 0.121 mol of total acid per 100. mL of juice, what volume (in mL) of 0.0998 M NaOH would you use to reach the endpoint of the titration?

Answer 1

Hi Bob!

So this question is semi poorly worded if they did not specify if it was a citric acid (what lemon just is) or a monoprototic acid. If it is an intro chemistry class, they probably mean monoprotic but I'll do both. (Monoprotic means one H+ to donate, think HCl or HBr)

*If* we are assuming it a monoprotic acid, then for 1 mol of the acid, we need 1 mol of base

an equation for molarity is M=n/V

If we have a monoprotic acid (1:1) ratio, M₁V₁=M₂V₂

so for acid, we can calculate molarity, M, by dividing the given 0.121 moles of acid by the volume **note volume is given in ml but we need to do in LITERS. so the Molarity is 0.121 mols/0.1 L giving a molarity of 1.21

SO going to M₁V₁=M₂V_{2, we can plug in 1.21 for} M_{1 and 0.1 L for} V_{1. So for the left side of the equation, we have 0.121 =} M₂V₂

From the problem, we were given for NaOH has a Molarity (M) of 0.0998. Because we have Molarity, and only one variable left in the M₁V₁=M₂V_2,

0.121=.0998 V ₂ so because we only have one variable left, we can solve for the volume of base we were asked about

.121/.0998=1.212 Liters of NaOH

If we are assuming it is citric acid, which is triprotic, that means for every mol of acid, we need 3 mols of base to neutralize it.

So, if we know from the problem the NaOH has 0.0998 M, we can use M=n/v to adapt the answer from above

Regardless M₁V₁=M₂V_{2, we can plug in 1.21 for} M_{1 and 0.1 L for} V_{1. (which we got from above) so for the left side of the equation, we have 0.121 =} M₂V₂

NOW THIS IS WHAT CHANGES

M=0.0998 mols / 1 Liter for the NaOH solution alone. But, we need 3 times the amount of mols that we have of the acid not the base. So, we take the given 0.121 mols of acid and multiply it by three giving us 0.363 mols of NaOH needed

We know the Molarity of the base solution is 0.0998. So if it now has 0.363 mols to keep the molarity the same we need to solve

.0998 M =0.363 mols / unknown liters solving this, we get 3.637 liters.

Finally, we go back to we want to keep the _{0.121 =} M₂V_{2 and plug in to check that each side of the equation is the same, meaning we did it right.}

_{0.121 = .0998 * 3.637.... giving 0.121=0.3629726}

_{what the heck?! that isn't equal... but it is RIGHT! Because of the 1:3 ratio, we can thinking of} M₁V₁=M₂V_{2 as} M₁V₁= 3 * M₂V_2,

Therefore, the final answer is 3.637 liters or 3637 mL (if we assume it is citric acid)

Given the bad wording, either 1.212 Liters of NaOH or 3.637 liters or 3637 mL would be correct.