pls can u solve this You perform an acid-base titration to standardize an HCI so

Question

pls can u solve this

You perform an acid-base titration to standardize an HCI solution by Placing 75.00 mL of HCl (aq) in a flask with a few drops of indicator solution. You put 0.1139 M NaOH(aq) into the burette and the initial reading in 151 mL. At the end point (i.e., when the solution just turns faint pink), the burette rending in to 39.11 mL. What is the concentration of the HCI (aq) solution? What volume of 0.0998 M Ba(OH)_2 solution would neutralize 50.00 mL of the HCl(aq) solution standardized in the problem 1, above? Sodium hydrogen carbonate, NaHCO_3 (sometimes referred to as bicarbonate) is often sprayed over spilled acid solutions. Write a chemical equation and net tome equation for the reaction that would take place.

Explanation / Answer

HCl + NaOH = NaCl +H2O

First calculate the volume of NaOH

39.11-1.51 mL = 37.6 mL=0.0376 L

Now number of moles of NaOH:

Molarity = number of moles / volume in L

Number of moles = 0.1139 M *0.0376 L

= 4.28*10^-3 mol NaOH

Now calculate the number of moles of            HCl as follows:

4.28*10^-3 mol NaOH*1.00 mol HCl/1.0 mol NaOH

= 4.28*10^-3 mol HCl

75.0 mL = 0.075 L

Molarity of HCl = 4.28*10^-3 mol HCl/ 0.075 L

= 0.057 M HCl

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