Explanation:
mole of acid/mole of base=(concentration of acid×volume of acid)/(concentration of base×volume of base)
Na/Nb=CaVa/CbVb
1/2=(Ca×16.7)/(0.2×25)
0.2×25=2×Ca×16.7
Ca=(0.2×25)/(2×16.7)
Ca=0.15mol/dm³
mole of acid=Concentration×volume
mole of acid=0.15×(16.7/1000)
mole of acid=2.5×10^-3moles
The answer is the concentration of H₂SO₄ = 0.25 mol/L or 0.25 M.
A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water.
The neutralization of a strong acid and strong base has a pH equal to 7.
The neutralization of a strong acid and weak base will have a pH of less than 7, and conversely, the resulting pH when a strong base neutralizes a weak acid will be greater than 7.
When a solution is neutralized, it means that salts are formed from equal weights of acid and base
H₂SO₄(l) + 2NaOH(l) - - - > Na₂SO₄(aq) + 2H₂O((l)
From the Equation
Mole ratio of H₂SO₄ and NaOH = 1 : 2
Molarity of Sodium Hydroxide = 0.200mol/dm3
Molarity = moles of solute/litre of solution
moles of solute = 0.200 * 25 *10⁻³
moles of NaOH = 5 *10⁻³ moles
As Mole ratio of H₂SO₄ and NaOH = 1 : 2
moles of H₂SO₄ = 2.5 *10⁻³
So, the concentration of H₂SO₄ = 0.25 mol/L or 0.25 M.
To know more about Neutralization reaction
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