Question

# HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) 4.00 mL of an unknown...

HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

4.00 mL of an unknown acid solution containing HCl was added to flask containing 30.00 mL of deionized water and two drops of an indicator. The solution was mixed and then titrated with NaOH until the end point was reached. Use the following titration data to determine the mass percent of HCl in the acid solution.

Mass of flask + acid solution: 129.50 g

Initial volume of NaOH from buret: 0.15 mL

Final volume of NaOH from buret: 16.70 mL

Molarity of NaOH: 0.5458 M

Given that Mass of flask: 125.59 g

Mass of flask + acid solution: 129.50 g

Mass of acid solution = 129.50g-125.59 g

= 3.91 g

HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

Volume of NaOH used = 16.70-0.15 = 16.55 ml = 0.01655 L

Moles of NaOH = Molarity * volume in L

= 0.5458 M * 0.01655 L

= 0.009 Moles NaOH

First calculate the moels of HCL in the solution as follows:

0.009 Moles NaOH *1/1

= = 0.009 Moles HCl

Mass of HCl = number of moles * molar mass

= 0.009 Moles HCl*36.46 g/mol

= 0.33 g HCl

% of HCL in the sample =[ 0.33 g HCl/3.91 g ]*100

= 8.44 %

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