Question

How many grams of the titrant used in the question 3 (sodium hydroxide -- 0.1001 mole/kg of solution) are required to titrate 10.00 g of a 5% (w/w) solution of butanedioic acid to the endpoint?

Answer 1

846.6 g of sodium hydroxide solution

Explanation:

The chemical reaction of butanedioic acid with sodium hydroxide:

HOOC-CH₂-CH₂-COOH + 2 NaOH → NaOOC-CH₂-CH₂-COONa + 2 H₂O

To find the mass of the solute (butanedioic acid) from a solution about which we know that the weight/weight percent concentration, we use the following formula:

concentration / 100 = solute mass / solution mass

solute mass = (concentration × solution mass) / 100

solute mass (butanedioic acid mass) = (5 × 10) / 100 = 0.5 g

number of moles = mass / molar weight

number of moles of butanedioic acid = 0.5 / 118 = 0.04237 moles

Knowing the chemical reaction we devise the following reasoning:

if        1 mole of butanoic acid reacts with 2 moles of sodium hydroxide

then   0.04237 moles of butanoic acid reacts with X moles of sodium hydroxide

X = (0.04237 × 2) / 1 = 0.08474 moles of sodium hydroxide

If the concentration of the titrant, sodium hydroxide solution, is 0.1001 moles / 1000 g of solution the we devise the following reasoning:

if there are         0.1001 moles of sodium hydroxide in 1000 g of solution

then there are   0.08474 moles of sodium hydroxide in Y g of solution

Y = (0.08474 × 1000) / 0.1001 = 846.6 g of solution

Learn more about:

weight/weight percent concentration

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