Question

QUESTION 4 4 points (Extra Credit) Save Answer Bonus question: Kemmi Major is working on a professor's research project. She uses a back-titration technique to determine if her reaction has formed the maximum amount of product. Back-titration works by reacting the species of interest (S) with an excess of the first chemical (A), resulting in complete reaction of the species of interest and some left-over first chemical. Then a second chemical (B) is titrated into the mixture to react with the remaining amount of the first chemical (A). Finally, the amount of the species of interest is determined by taking the total mole amount of the first chemical and subtracting the mole amount of the second chemical (for the simplest case of one-to-one mole ratios). Put into a formula: total moles of A = total moles of S + total moles of B This re-arranges to form total moles S = total moles A-total moles B First Kemmi takes just 5.00 ml. out of her reaction mixture. Next, she quenches the reaction (quenching stops the reaction abruptly) by adding exactly 10.00 ml of 0.1500 M hydrochloric acid. The hydrochloric acid reacts with Kemmi's product of interest. Then Kemmi titrates the mixture with 3.24 mL of 0.1500 M sodium hydroxide. Calculate the moles of hydrochloric acid added (using the volume and concentration of the acid) Calculate the moles of sodium hydroxide titrated into the mixture (using volume and concentration of the base) All the mole ratios are 1:1, so the formulas given above are used. Determine the mole amount of Kemmi's product by subtracting the moles of sodium hydroxide from the moles of hydrochloric acid. Finally, calculate the concentration of Kemmi's product in mol/L (Remember, the moles of product were originally in 5.00 mL of solution.) Enter the answer with 4 significant figures and units of mol/L

Answer 1

Ans. The no. of moles of HCl(A)added$\rightarrow$

volume (in ml)*concentration = 10ml*0.15M=1.5 milli moles

The no. Of moles of NaOH(B) added$\rightarrow$

volume(in ml)*concentration = 3.24*0.15 = 0.486 millimoles

Moles of S = Moles of A - Moles of B = 1.5 - 0.486= 1.014 millimoles.

Concentration of S = 1.014*10^-3mol/5*10^-3litre=0.2028mol/L

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