Question

calculate the volume of 0.5N Hcl solution required to neutralize 25ml of 0.1N Na2Co3 solution ​

Answer 1

Answer:

The reaction for neutralizing sodium carbonate with hydrochloric acid should go as follows:

Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)

For neutralization we can use the equation:

V1 • N1 • n2 = V2 • N2 • n1, where V is volume (mL) N is the concentration (N), and n is molar equivalents for the reaction.

Given what we know, the hydrochloric acid has a concentration of 0.5 N (N1) and a molar equivalent of 2 for this reaction (n1), whereas sodium carbonate has a concentration of 0.25 N (N2), a volume of 25 mL (V2) and a molar equivalent of 1 (n2). Therefore:

V1 • 0.5 N • 1 = 25 mL • 0.25 N • 2

V1 = (25•0.25•2)/(0.5•1) = 12.5/0.5 = 25 mL

So 25 mL of 0.5 N HCl should be required

Explanation:

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