47.0 mL of a H2CO3 solution were titrated with 37.5 mL of a 0.215 M LiOH solution to reach the equivalence point, what is the molarity of the H2CO3 solution?

Question

I cant seem to get the correct answer, please help!

Answer 1

Same deal here - the equivalence point is reached when you add enough OH- to react with all of your H+.

 

The balanced equation for this reaction is:

 

H₂CO₃ + 2LiOH → Li₂CO₃ + 2H₂O

 

I showed you the long (but safe!) way to calculate the concentration of the H3PO4 on that other one.  Now I'll show you the "shortcut" way.  ;)

 

Since the moles of acid (H+) must equal the moles of base (OH-), we can look at how to find those and set them equal to each other.  Isn't the number of moles of H+ provided by the acid equal to the moles of acid times how many H's it has?  Therefore, the moles of H+ should equal:

 

(Volume of acid) • (molarity of acid) • (number of H's per mole of acid) = moles of H+

 

Same thing for the base, the moles of OH- should equal:

 

(Volume of base) • (molarity of base) • (number of OH's per mole of base) = moles of OH-

The formula to relate these uses "n" for the number of H+ or OH- per mole of acid or base, V = volume, and M = molarity.

 

n_AM_AV_A = n_BM_BV_B 

 

If you do a titration, figure out what "n" is for the acid and the base by looking at the formulas. H₂CO₃ has 2 H+ on it, so n_A = 2.  LiOH has one OH- on it, so n_B = 1. The question says that 47.0 mL of a H2CO3 solution were titrated with 37.5 mL of a 0.215 M LiOH.  The only missing thing is the concentration of H2CO3. And the cool thing about the formula is that you can use the volume in mL or L, as long as they are both the same they will cancel out!  ;)

 

Rearrange the formula to solve for what you are missing (molarity of the acid) and plug in the known values.

 

M_A = n_BM_BV_B / n_AV_A  = (1)(0.215 M)(37.5 mL) / (2)(47.0 mL) = 0.0856 M H₂CO₃    :)

Answer 2

We know that the moles of H2C03 in 47 mL = the moles in 37.5 mL of 0.215M LiOH as it is at the equivalence point.

Thus 0.047 L * x mol/L = 0.0375 L * 0.215 mol/L

or x = (0.0375 * 0.215) mol / 0.0470 L = 0.172 M to three significant figures.