Chemistry problem?? I need someone to help me solve this chemistry work for me?

Question

HCI reacted with 0.100 moldm-3 NaOH in a titration using methyl orange as indicator. 20.90cm3 of the acid of unknown concentration neutralized completely 25.00cm3 of the base.

(i) calculate the concentration of the acid in moldm-3.

(ii) if the acid was prepared by dissolving 5.0g of HCI in 1dm3 of solution, calculate the percentage purity of the acid

If 100cm3 of the acid was diluted to 500cm3,calculate the

(i)dilution factor

(ii)concentration of the new solution in moldm-3

Please show detailed working for me. 5 stars and best answer for anyone who shows detailed explanation.

Answer 1

sorry confusing

Answer 2

First off, moldm^-3 is the same as moles/L, or molarity (M). Also, cm^3 = mL.

HCl + NaOH ==>  H2O + NaCl

HCl and NaOH react in a 1:1 mole ratio, so moles NaOH = moles HCl.

moles NaOH = M NaOH x L NaOH = (0.100)(0.02500) = 0.00250 moles NaOH = moles HCl

M HCl = moles HCl / L = 0.00250 / 0.02090 = 0.120 M HCl

5 g HCl / L x (1 mole HCl / 36.5 g HCl) = 0.137 M. But our actual M = 0.120 M, so the acid must be (0.120 / 0.137) x 100% = 87.6% pure

If 100 mL of the acid is diluted to 500 mL, the dilution factor is (500 / 100) = 5.

To find the M of the new solution,

M1V1 = M2V2

M1 = M of original acid = 0.120

V1 = mL of original acid = 100

M2 = M of diluted acid = ?

V2 = mL of diluted acid = 500

(0.120)(100) = M2 x (500)

0.0240 M = M2

Answer 3

First, write a balanced chemical equation describing the reaction:

HCl + NaOH -> NaCl + H2O

Next, determine the amount of base which reacted with the acid. From the equation above, the molar ratio of acid to base is 1 to 1. The moles of base is (25 mls)*(1 liter/1000 mls)*(0.100 moles/liter) = 0.0025 moles. This also means 0.0025 moles of acid were used. The concentration of the acid was (0.0025 moles)/(20.90 mls)*(1 liter/1000 mls) = 0.1196 moles/liter.

To calculate the purity, find out how many moles 5 grams of HCl represents. This would be the weight divided by the molecular weight. 5 grams / (36 grams/mole) = 0.138 moles HCl. If this was dissolved in 1 liter, then the calculated molarity was 0.138 moles/liter. The titration revealed the true molarity was only 0.1196 moles/liter. This means the purity is 0.1196/0.138 = 86.6%

100 mls diluted to 500 mls is a dilution factor of 500/100 = 5. The new concentration would be the number of moles over the new volume. (0.1196 moles/liter)*(100 ml)*(1 liter/1000 mls)(/0.5 liters) =  0.0239 moles/liter.