Question

what is the concentration of the naoh(aq) given that 20.8cm3 of 0.0500 moldm-3 h2so4 neutralises 25.0cm3 of it?

Answer 1

Answer on Question #63561, Chemistry / Inorganic Chemistry

Question:

what is the concentration of the naoh(aq) given that 20.8cm³ of 0.0500 moldm-3 h2so4 neutralises 25.0cm³ of it?

Solution:

1) Write down the balanced equation:

$$2 \mathrm{NaOH} + \mathrm{H_2SO_4} = \mathrm{Na_2SO_4} + \mathrm{H_2O}$$

We can see that 1 mole of acid neutralizes 2 moles of NaOH.

2) Let's calculate amount of moles of acid in solution:

$$20.8 \mathrm{cm^3} = 0.0208 \mathrm{dm^3}.$$

So $0.0208 \mathrm{dm^3}$ of $0.0500 \mathrm{moldm^{-3}}$ solution contains $0.0208 * 0.0500 = 0.00104$ moles of acid.

3) Using 1) we can conclude that NaOH solution contains $0.00104 * 2 = 0.00208$ moles of NaOH.

$25.0 \mathrm{cm^3} = 0.0250 \mathrm{dm^3}$. So the concentration is $0.00208$ moles / $0.0250 \mathrm{dm^3} = 0.0832$ mol/dm³.

Answer:

The concentration of NaOH solution is $0.0832 \mathrm{mol/dm^3}$.

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