Question

50cm3 of sodium hydroxide solution added to a solution of sulfuric acid. The concentration of the sodium hydroxide solution was 20g/dm3. Work out the concentration of the sulfuric acid in grams per dm3 if it took 25cm3 of it to completely neutralise the sodium hydroxide. The relative molecular mass of sulfuric acid is 98.

Answer 1

Step-by-step explanation:

first you need to know what's the reaction that's taking place

h2so4 + 2 naoh -> na2so4 + 2 h2o

from this equation you know that 1mol h2so4 neutralises 2 mol naoh, so there's a relation that you can use as follows

`(1 \: mol \: h2so4)/(2 \: mol \: naoh) = (h2so4 \: molarity \: * vol)/(naoh \: molarity \: * vol)`

rearranging the terms you get

`(naoh \: molarity \: * vol)/(2 \: mol \: naoh) = (h2so4 \: molarity \: * \: vol)/(1 \: mol \: h2so4)`

naoh molarity is

20g/L × 1mol/40g = 0.5 mol / L

using the data you have, in the formula

`\frac{0.5mol * {l}^( - 1) * 50 * {10}^( - 3) l }{2 \: mol \: naoh} = \frac{h2so4 \: molarity * 25 * {10}^( - 3) l}{mol \: h2so4}`

you get that

h2so4 molarity = 0.5 mol/L

if h2so4 molar mass is 98g/mol, then

(0.5mol / L) × (98g/mol) = 49g/L

then the concentration of the sulfuric acid in grams per dm3 is 49g/dm³