Calculate the molar amounts of NaOH used in the reaction with the HCL solution a

Question

Calculate the molar amounts of NaOH used in the reaction with the HCL solution and with the HC2H3O2 solution. AND Calculate the molar concentration(molarity) of the HCL solution and the HC2H3O2 solution. we started with 20 ML of distilled waterand 5 ML of HCL solution into the 100 ML beaker containing both. (a) rinse the reagent reservoir with a few mL of the 0.100 M NaOH solution. (b) fill the reagent reservoir with slightly more than 60mL of the 0.100 M NaOH solution. (c) place a 250 mL beaker, which contains the NaOH rinse beneath the tip of the reservoir. (d) drain a small amount of the NaOH solution into the 250 mL beaker so that it fills the reservoir's tip. We let the process go on until be reached the titration. Then "Repeat the titration with an acetic acid, CH3COOH, solution of unknown molar concentration. Do this in a manner of the first titration. I'm not sure if this is all the information you need? I'm not really sure where to start.

Explanation / Answer

1 molar quantiy of HCL will require 1 molar quantity of NaOH reason both are storng acids and bases respectively with a single equivalent. So HCl + NaOH = NaCl +H2O Molar weight of HCl = 35.5 10 ml of HCL of molarity 0.1014 = 0.1014 * 35.4 /100 = 0.036 gms in 60 ml = 0.05 m approx so you would require 0.05 molar solution of 60 ml Molarity is in mol/L. You have: 20.00 mL of acetic acid*(1.049g/ml)=20.89 g acetic acid The molecular weight of acetic acid is 60.05 g/mol from the periodic table. Use this information to convert to moles of acid. 20.89 g acetic acid * (1mol/60.05g)=0.349mol You need the TOTAL volume of the solution: 20ml+250ml=270ml and convert that to L: 0.270L Now to get the molarity, divide moles/L: 0.349mol/0.270L=1.29 M solution

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