What volume of 0.600m h2so4 would be required to reach the equivalence point in a titration with a solution containing 2.50g of nahco3

Hey there!:

Reaction stoichiometry :


Number of moles NaHCO3 :

Molar mass NaHCO3 =  84.007 g/mol

n = m / mm

n = 2.50 / 84.007

n  = 0.0297moles of NaHCO3

 2 NaHCO3 + H2SO4 = Na2SO4 + 2 CO2 + 2 H2O

2 moles NaHCO3 1 mole H2SO4
0.0297 moles NaHCO3 moles H2SO4 

moles H2SO4 = 0.0297 * 1  / 2

moles H2SO4 = 0.0297 / 2

=  0.01485 moles of H2SO4

Therefore:

Molarity ( H2SO4 )  =  moles H2SO4 /  volume

0.600 M = 0.01485 / V

V = 0.01485 / 0.600

V = 0.02475 L of H2SO4

hope this helps!

C

Explanation:

hydrogen Bond exists between ammonia, hydrogen flouride and water so the answer is C