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aleksklad [387]
3 years ago
10

Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume

d during the timed reaction? Calculate the number of moles of H2O2 present initially and determine how many have reacted with iodide when the thiosulfate runs out. How does the I- concentration change during this same time?
Chemistry
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

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If 50.0 g of formic acid (hcho2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of soluti
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If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.

Therefore, option C is the correct option.

Given,

Given mass of sodium formate = 30 g

Given mass of formic acid = 50 g

Volume of sodium formate = 500 ml

Volume of formic acid = 500ml

Molar mass of sodium formate = 68 g

Molar mass of formic acid = 46 g

<h3>To calculate concentration of sodium formate and formic acid</h3>

The ratio of number of moles and the volume of solution is Molar concentration of substance.

Concentration of sodium formate

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= 0.00088m

Concentration of formic acid

Ca = 50/(46×500)

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Now,

by using Henderson hesselbalch equation,

pH = pKa + log(Cb/Ca)

pKa = -log(1.8 × 10^(-4))

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pH = 3.75 + log(0.00088/0.00217)

pH = 3.75 - 0.392

pH = 3.35

Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.

learn more about pH:

brainly.com/question/13423434

#SPJ4

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