0.30 g of citric acid is dissolved in distilled water to produce 100.0 mL of a 0.016 M solution. How much citric acid is needed in order to complete a full neutralization process with 30.0 ml of...

0.30 g of citric acid is dissolved in distilled water to produce 100.0 mL of a 0.016 M solution. How much citric acid is needed in order to complete a full neutralization process with 30.0 ml of 0.50 M sodium bicarbonate solution? 

Expert Answers
ncchemist eNotes educator| Certified Educator

Citric acid has three acidic protons that can be neutralized by the sodium bicarbonate.  First we need to find the number of moles present of sodium bicarbonate and then use that to determine the number of moles of citric acid required.  Finally we will convert that number of moles of citric acid into the volume of the starting solution that corresponds to.

First we start with the sodium bicarbonate solution.  We will use 30 mL of a 0.5 M solution of NaHCO3.  Let's convert that into the number of moles of NaHCO3 by multiplying the concentration by the volume:

30 mL * (1 L / 1000 mL) * (0.5 moles / L) = 0.015 moles NaHCO3

Since 3 moles of NaHCO3 are required to neutralize one mole of citric acid (remember that citric acid has 3 acidic protons), that means the 0.015 / 3 = 0.005 moles of citric acid are required for a full neutralization of the given NaHCO3.  Now we need to convert 0.005 moles of citric acid into the volume of the 0.016 M solution of citric acid required to achieve that number of moles:

0.005 moles * (1 L / 0.016 moles) =  0.3125 L of citric acid solution

So we would need 0.3125 L (or 312.5 mL) of the citric acid solution to complete the neutralization.

jerichorayel | Student

The balanced chemical reaction for this problem is:

 

`C_6 H_8 O_7 + 3 NaHCO_3 -> 3 CO_2 + 3 H_2O + Na_3C_6H_5O_7 `

First, we need to get how many moles of sodium bicarbonate to be neutralized. 

Moles of NaHCO3 = 

 

`30 mL * ((1L)/(1000 mL)) * ((0.5 mol es NaHCO_3)/L) = 0.015 mol es NaHCO_3 `

 

To get the moles of citric acid (C6H8O7) that will react with 0.15 moles of NaHCO3 = 

 

`0.015 mol es NaHCO_3 * ((1mol e C_6H_8O_7)/(3 mol es NaHCO_3)) = 0.0050 mol es C_6H_8O_7`

Now we can solve for the amount of citric acid needed to completely react with sodium bicarbonate.

(1) In terms of mass:

 

`0.005 mol es C_6H_8O_7 * ((192.12 grams)/ (1 mol e C_6H_8O_7)) = 0.96 grams of C_6H_8O_7`

 

(2) In terms of the citric acid solution:

Molarity = moles/volume of solution

Volume of solution = moles citric acid/molarity of the citric acid solution

`Volume of solution = (0.005/0.016) = 0.31 L`