How many grams of NaHCO3 should be added to one liter of 0.100 M H2CO3 (Ka = 4.2 x 10-7) to prepare a buffer with pH = 7.00?

Question

I got molefrom.100/1L equal .100mole. where do I go from here?

Answer 1

1st:7= -log( 4.2x10^-7)+log(NaHCO3÷.100)
2nd: 7=6.3767+log(NaHCO3÷.100)
3rd: minus6.3767 on both side which equal .6232 lead to -->.6232= log(NaHCO3÷.100)
4th: 10^.6232=10^log(NaHCO3÷100)
5th: 4.2= (NaHCO3÷100)
6th: cross multiply:4.2x.100=4.2mole now NaHCO3=4.2mole
7th: multiply molar mass: 4.2x94g/mole=?
answer: 39gram

Answer 2

There are a mistake in the calculation. The correct calculation
4.2x.100=.42mole now NaHCO3=.42mole
7th: multiply molar mass: .42x84g/mole=?
answer: 35gram not39gram