Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1494 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 18.48 mL of the standard solution. What is the concentration of the original sulfuric acid solution? concentration: 2.76

Answer 1

Concentration = 1.38 M

- Na2SO, lar) + 2 H2O H) H₂SO4 (aq) + 2NaOH (ay) _(1 mase) __ [2 male) - - According to balanced chemical equation neutralise I mole of sulphuric acide 2 mole Nach we know that the volume of NaOH used = 18.48 ml. and concentration of Naon_soliclions 0-14.94 M which means that 0.1494 males of NaOH are present in 1000 me /1 litre solution je 1000 ml solution contains - 0.1494 moles of NaOH. then

1 me of solution contain 0.1494 1000 moles of NaOH I and. ' 18.48 me of solution contain 0.1494 X 18.48 males 1000 _ 0.00 2760912 males ...0.002760912 moles of NaOH are used to neutralise _10_me diluted solution of H2SO4.. je 10 me solution of H2SO4 required 0.002760 912 moles. Ime solution of H2S0y_crequired 0.002760912 moles 10 and, 250 ml solution of H2S0y required 0.002760912 x 250 10 + 0.0690228 mells of H y NaOH

0.0690 228 moles of NaOH are required to neutralise 250 me diluted sample of H2SO4.. --- -OB- 6.0 690228 moles of NaOH are required to neutralise 25 me original sample of H2SO I because moles of H2SO4 uzemains conserved on dilutione _ -_ - --- According to balanced chemical equation 2 mole NaOH required 1 mole H2SO4 I more NaOH required - 1 mole H2SO4 and 0.0690228 moles Naot required 1 x 0.0690228 2 0:0 34 5114_moles of H2S Oy so molarity 3 miles of H2S0y Volume of H2SO4 solution (litre) _ = 0.0345114 malesa __ 0.025 L 1.380456_M lar, concentration = 1.38 / M.