One student found that 25.0 ml of sodium hydroxide solution was neutralized by e

Question

One student found that 25.0 ml of sodium hydroxide solution was neutralized by exactly 22.30 ml of 0.100 M sulfuric acid solution. (2 mol of NaOH + 1 mol of H_2 SO_4(aq) rightwardsdoublearrow 1 mol of Na_2 SO_4 + 2 mol of 2H_2 O_(l)) Use formula M_A Y_A/n_A = M_B Y_B/n_B (i) Calculate the number of moles of sulfuric acid. number of moles = ____________ (ii) Calculate the concentration, (in Molarity) of the sodium hydroxide solution used. Give your answer to 3 decimal places. Concentration = _______________ M If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H2SO4), what is the concentration of the H2SO4 solution? Can U titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences. Explain the difference between an endpoint and equivalence point in a titration.

Explanation / Answer

Ans 2. (i)

Molarity = no. of moles of H2SO4 / volume of Solution in Litre

0.100 = n / 0.0223

n = 0.00223 moles

Number of moles of sulfuric acid = 0.00223

Ans 2 (ii)

Now 1 mole of H2SO4 requires 2 moles of NaOH

So 0.00223 moles will require ( 2x 0.00223) = 0.00446 moles

Using the formula

MaVa / na = MbVb / nb

Where M is the molarity , V is volume and n is the number of moles

(Ma x 25) / 0.00446 = (0.100 x 22.30) / 0.00223

Ma = 0.1784 M

So the concentration of NaOH will be 0.178 M

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