A 250.0-mL solution of H2SO4 was analysed by taking a 100.0-mL portion (aliquot)

Question

A 250.0-mL solution of H2SO4 was analysed by taking a 100.0-mL portion (aliquot) and adding 50.00 mL of 0.213 M NaOH (aq). After the reaction occurred, an excess of OH^- ions remained in the solution. The excess base required 13.21 mL of 0.103 M HCI(aq) for neutralisation. Calculate the molarity of the original solution of H_2SO_4. A sample may contain any or all of the following ions: Hg_2^2+, Ba^2+ and Mn^2+. No precipitate formed when an aqueous solution of NaCl or Na_2SO_4 was added to the sample solution. A precipitate formed when the sample solution was made basic with NaOH. Which ion or ions are present in the sample solution?

Explanation / Answer

Between HCl and NaOH
HCl + NaOH -----> NaCl + H2O
M1V1 = M2V2
0.213 * V1 = 0.103 * 13.21
V1 = 6.3879 mL
Between H2SO4 and NaOH
H2SO4 + 2 NaOH ------> Na2SO4 + 2 H2O
M1V1/n1 = M3V3/n3
0.213*(50-6.3879)/2 = M3*100/1
M3 = 0.04644 M

if Ba+2 is there then it forms BaSO4 precipitate with Na2SO4.
and Hg(OH)2 can not exist in aqueous state.
Henace the precipitate is formed by Mn+2 and it is Mn(OH)2

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