Question

30.0cm3 of 1.52 M sodium hydroxide solution is added to 20.0cm3 of 1.14 M sulphuric acid. What is the molarit of the resulthing sodium sulphate solution?

Answer 1

@shubhamsrg would you kindly help me??

Answer 2

final molarity =moles of sodium sulphate/total volume(in litres)

any attempt ?

Answer 3

wait...

Answer 4

i couldn't get it..

Answer 5

write the balanced chemical reaction of NaOH and H2SO4

Answer 6

\[2NaOH+H_{2} SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O\]

Answer 7

correct
find of the respective moles of NaOH and H2SO4 from the given information.
Use M = moles/vol , convert the given volume in litres.

Answer 8

mole of NaOH=0.0456mol
\[mole of H_{2}SO_{4}=0.0228mol\]

Answer 9

volume in dm3

Answer 10

that is correct again

now note that 2 moles of NaOH react with 1 mole of H2SO4 to give 1 mole of Na2SO4

hence, 0.0228 moles of H2SO4 will react with 0.0456 moles of NaOH to give 0.0228 moles of Na2SO4
following till now ?

Answer 11

yup...

Answer 12

then your required answer will be
M= moles of Na2SO4/total volume

total volume = 50cm3 here, convert that into litres

Answer 13

0.456M?

Answer 14

yep.

Answer 15

thanks

Answer 16

glad to help

Answer 17

and i have many questions><