Answer to Question #287823 in General Chemistry for elle

Question

if 25cm³ of a soltuion of NaOH are neutralised by 27.5cm³ of H2SO4 of a concentration of 0.250moldm^-3 what is the concentration of the NaOH?

Answer 1

Balanced chemical equation for the reaction:

2NaOH + H₂SO₄ products Na₂SO₄ + 2H₂O

Provided data:

NaOH; volume - 25cm³ = 0.025dm³

H₂SO₄; volume - 27.5cm³ = 0.0275dm³

concentration - 0.250moldm^-3

• Applying the formula C=n/V;where n is moles,C is concentration and V is Volume,we then calculate the moles of 0.0275dm³ H₂SO₄ that reacted with NaOH.
• n=CV

n (moles of H₂SO₄) = 0.250mol/dm³ H₂SO₄ × 0.0275dm³ H₂SO₄ (dm³ cancels out).

= 6.875 × 10^-3 moles H₂SO₄

From the balanced equation,

Mole ratio of NaOH : H₂SO₄ is 2:1

Therefore,

6.875 × 10^-3 moles H₂SO₄ × 2 moles NaOH/1 moles H₂SO₄

 

= 1.375 × 10^-2 moles NaOH

• Applying the formula C=n/V, we calculate the concentration of NaOH.

n=1.375 × 10^-2 moles, V= 0.025 dm³

C= 1.375 × 10^-2 moles/0.025 dm³ = 0.55 moldm^-3

Therefore, concentration of NaOH is 0.55 moldm^-3