chem question answer help

Question

A volume of 11.0 mL of a 0.32 M solution of hydrogen phosphate (or phosphoric acid) reacts completely with 71.0 milliliters of sodium hydroxide. What is the molarity of the sodium hydroxide solution?

Answer 1

Hello, Molly,

This is a bit difficult (for me) to answer. I'll assume, as you confirm, that hydrogen hosphate is phosphoric acid. The term is sometimes used as shorthand for other forms of the phosphate group, such as sodium hydrogen phosphate.

H₃PO₄ (phoshoric acid) has three hydrogen aroms that can dissociate and react with the sodium hydroxide. The titration curve actually has three values of pH (or the amount of base, NaOH here) corresponding to each of these hydrogens. The base must be strong enough to remove the last hydrogen.

Since none of the equilibrium coefficients are provided for the three hydrogens, I'll asume that the base is strong enough to remove all three, and that the endpoint of the titration is set at a pH high enough to measure the point at which all three are removed.

The chemical equation tells us we need three times as much NaOH as phosphoric acid to reach complete neutralization, where all three hydrogens are removed:

H₃PO₄ + 3NaOH = Na₃PO₄ + 3H₂O

Three moles of NaOH to neutralize one mole of phosporic acid. If you are studying Normality in addition to Molarity, the use of N as a concentration unit takes into account the multiplicity of H's on phosphoric acid. But I'll do the calculation based on moles.

If the molar ratio of an acid/base titration was 1:1, we could use the expression

M1V1 = M2V2,

where M is concentration (M) and V is volume (the unit for volume can be anything, as long as it is the same on both sides. The subscripts 1 and 2 represent the two components beting titrated.

Make a table, where 1 is the phophoric acid and 2 represents the sodium hydroxide:

M1 = 0.32M

V1 = 11 ml

M2 = ? - the unknown

V2 = 71.0 ml

If the reaction were 1:1, we would simply find the solution (concentration of NaOH) as follows:

M1V1=M2V2

M2=M1V1/V2

Let's go ahead and calculate M2 based on a 1:1 molar ratio. We can multiply the value by 3 afterwards to obtain the value for our 1:3 molar ratio (phosphoric acid/NaOH) titration.

M2 = (0.32)*(11ml)/(71.0ml)

M2= 0.0496 M

We need to multiply this by 3 to account for the molar ratio:

= 0.149M

There arfe actually only 2 sig figs, so this becomes 0.15 M NaOH.

I hope this helps, and that I am correct,

Bob