Question

o
A solution contains Na,co, and NaHCO3. 10 mL of the solution required
2.5 mL of 0.1MH,So, for neutralisation using phenolphthalein as
indicator. Methyl orange. is then added when a further 2.5 mL of 0.2 M
H,SO, was required. The amount of Na,Co, and NaHCO, in 1 litre of the
solution will be:
amount of Na, Co, is 5.3g
amount of NaHCO, is 0.042 g
amount of Na,Co, is 0.053g
amount of NaHCO, 4.2g​

Answer 1

Answer:

Explanation:

equivalent of na2co3  = equivalent of acid

moles of na2co3*1000=0.5

moles of na2co3 = 5*10^-4

equivalent of nahco3 formed + equivalent of nahco3 initial = 1

moles of nahco3 + moles of nahco3 = 0.001

moles of nahco3 initial = 10^-3 - 5*10^-4

=5*10^-4

molarity of na2co3 = 5*10^-4 / 0.01

molarity of na2co3 = 5*10^-2

molarity of nahco3 = 5*10^-2

moles of na2co3 in 1lt solution = 5*10^-2

weight = 5gm

moles of nahco3 in 1lt solution = 5*10^-2

weight = 4.2 gm

so option d is correct