Question

If 36.10 mL of 0.223 M of NaOH is used to neutralize a 0.515 g sample of citric acid, what is the molar mass of the acid?

Answer 1

The molar mass of the citric acid is `"192 g/mol"`.

The balanced chemical equation for the neutralization reaction that takes place between sodium hydroxide, `NaOH`, and citric acid, `C_6H_8O_7`, looks like this

`C_6H_8O_(7(aq)) + color(red)(3)NaOH_((aq)) -> Na_3C_6H_5O_(7(aq)) + 3H_2O_((l))`

Notice the `1:color(red)(3)` mole ratio that exists between citric acid and sodium hydroxide; what this tells you is that, for every mole of citric acid, you need 3 times more moles of sodium hydroxide for the reaction to take place.

Since you know the molarity and the volume of the `NaOH` you've used, you can calculate how many moles of `NaOH` reacted

`C = n/V => n = C * V`

`n_(NaOH) = "0.223 M" * 36.10 * 10^(-3)"L" = "0.00805 moles"`

Now use the aforementioned mole ratio to see how many moles of citric acid were present in 0.515 g

`0.00805cancel("moles NaOH") * "1 mole citric acid"/(3cancel("moles NaOH")) = "0.00268 moles citric acid"`

Now simply divide the mass of citric acid given by the number of moles it contained to get the compound's molar mass

`M_M = m/n = "0.515 g"/("0.00268 moles") = "192.16 g/mol"`

Rounded to three sig figs, the number of sig figs given for 0.515 g, the answer will be

`M_M = color(green)("192 g/mol")`

SIDE NOTE The actual molar mass of citric acid is 192.12 g/mol, so your result is in agreement with the known value.