Question

A student added 32.60 mL of 0.03020 M NaOH to a 20.0 mL sample of ginger ale before his sample turned pink. (a) How many moles of NaOH did he add to his solution? (b) How many moles of citric acid are in his sample? (c) How many grams of citric acid does this correspond to? (d) How many milligrams? How many mg/mL?

Answer 1

Solution:

molarity= $\frac{\text{moles}}{\text{volume in litres}}$

(a) Molarity of NaOH, $M_1$= 0.03020 M

Volume of NaOH added,$V_1$= 32.60 mL= 0.0326 L

(1L= 1000mL)

Number of moles of NaOH added :

0.03020 M NaOH means 1L of solution contains 0.03020 moles 0.0326 L contains =$0.03020\times 0.0326$ = 0.00098 moles

(b) Equal number of moles of acid  will neutralize equal number of moles of base, thus we can apply molarity equation:

Molarity of citric acid, $M_2$=?

volume of citric acid, $V_2$= 20 mL= 0.02 L

using formula, $(M_1)(V_1)=(M_2)(V_2)$ ,

$M_2= \frac{(M_1)(V_1)}{(V_2)}$

$M_2= \frac{(0.03020 M)(0.0326 L)}{(0.02 L)}$ = 0.049 M

(c) Grams of citric acid in 20 ml

moles of citric acid = ${M_2}\times V_1$ of solution in Liters = $0.049 \times 0.02$ = 0.00098 moles

Mass of citric acid =$\text{moles}\times \text{Molecular} mass$

$= 0.00098\times 192.12$ = 0.1882 g

(d) 20 mL solution of ginger ale contains of citric acid

$0.1882 g= 0.1882 \times 1000 mg$ = 188.2 mg (1g = 1000mg)

Since, 20 mL solution of ginger ale contains of citric acid 188.2 mg

so 1 mL of solution contains = $\frac{188.2 mg}{20 mL}$

                                              = 9.41 mg/ mL

Answer 2

(a)

Molarity = moles of solute / liter of solution

Moles of solute = molarity x liter of solution  

Molarity of NaOH = 0.03020 M

Liter of solution = 32.60 mL / 1000 = 0.0326 L

Moles of NaOH = 0.03020 M x 0.0326 L = 9.85 x 10⁻⁴ moles

(b)

The reaction between NaOH and citric acid is as follows:

C₃H₅O(COOH)₃ + 3NaOH → Na₃C₃H₅O(COO)₃ + 3H₂O

As the balanced chemical equation 3 moles of NaOH reacts with 1 mole of citric acid (Na₃C₃H₅O(COO)₃)

Since the moles of NaOH is 9.85 x 10⁻⁴ the moles of citric acid will be (9.85 x 10⁻⁴) / 3 or  3.284 x 10⁻⁴ moles.

Therefore, the moles of citric acid is  3.284 x 10⁻⁴

(c)

Moles = Mass/ Molar mass

Mass = Moles x Molar mass

Moles of citric acid = 3.284 x 10⁻⁴

Molar mass of citric acid = 192.124 g/mol

Mass of citric acid = 3.284 x 10⁻⁴ x 192.124 g/mol =

Therefore, the mass of citric acid is 0.0631 g

(d)

Mass of citric acid = 0.0631 g

1 g = 1000 mg

Therefore, mass of citric acid = 0.0631 x 1000 = 63.1 mg

(e)

Given,

Volume of citric acid = 20.0 mL

20 ml of citric acid has a mass of 63.1 mg

Therefore, miligram per militier = 63.1 mg/20 ml = 3.155 mg/ml