Question

Sodium benzoate, the sodium salt of benzoic acid, is used as a food preservative. A sample containing solid sodium. benzoate mixed with sodium chloride is dissolved in mL of M HCl, giving an acidic solution (benzoic acid mixed with HCl). This mixture is then titrated with M NaOH. After the addition of mL of the NaOH solution, the pH is found to be . At this point, the addition of one more drop ( mL) of NaOH raises the pH to . Calculate the mass of sodium benzoate ( ) in the original sample. (Hint: At the equivalence point, the total number of moles of acid [here HCl] equals the total number of moles of base [here, both NaOH and NaC₆H₅COO])

Short answer

Moles of present in the original sample is 0.0067 mol .

Step-by-step solution

 Step 1: Given Chemical Equation

The reaction between sodium benzoate and hydrochloric acid is given below

C₆H₂COONa + HCL $\rightleftharpoons$C₆H₅COOH + NaCl

From stoichiometry, 1mol of C₆H₅COONa reacts with 1 mol of HCL to produce 1 mol of C₆H₅COOH and 1 mol of NaCl .

Now, according to the question. HCl is added in excess. This excess HCl gets titrated with NaoH .

The reaction is given below:

HCl + NaOH$\rightleftharpoons$NaCl +H₂O

Calculating Change in pH

According to the question, C₆H₅COONa is mixed with NaCl and dissolved in 50mL of 0.5M HCL , further titrated with 0.393M NaoH

Now calculate the amount of added as shown below:

50 mL of 5M HCL= $\frac{1mol\times 50mL\times 0.5M}{1000mL\times 1M}$

= 0.025mol HCl

Calculate the change in pH when a drop of is added as below:

Change in pH =9.3 - 8.2

= 1.1

This is the equivalence point. At the equivalence point

Moles of HCl = Moles of NaOH

The excess moles of HCl reacts with NaoH added.

Now calculate the amount of NaoH added as shown below

46.5mL of 0.393M NaOH =$\frac{1mol\times 46.5mL\times 0.393M}{1000mL\times 1M}$

= 0.0183mol NaOH

Calculating moles of HCl

Since at the equivalence point,

Moles of HCl = Moles of NaOH

Excess moles of HCl = 0.0183 mol

Calculate the moles of that react with sodium benzoate as shown below:

Moles of HCl = Moles of NaOH

Excess moles of HCl = 0.0183 mol

Since 0.0067 mol of HCl react with 0.0067mol of C₆H₅COONa .

Therefore, moles of C₆H₅COONa present in the original sample is 0.0067 mol .