in the mixture of [NaHCO3+Na2CO3] volume of hcl required is x ml with phenolphthalein indicator and then y ml with methyl orange indicator in same titration.The volume of Hcl for complete neutailization of NaHCO3 in the mixture is

[a] y [b] 2x [c] y-2x [d] x/2

Let x be the amount of na2co3 in mixtureNo. of moles of na2co3 = (x/106) = aNo. of moles of nahco3 = [(1-x)/84] = ba = bTherefore, x= 0.5579Then, a = b = 0.0053By reaction of hcl and na2co3 it is clear that per 1 mol of na2co3 2 mol of hcl is required, so for 0.0053 mol of na2co3 we need 0.0106 mol.By reaction of hcl and nahco3 it is clear that per 1 mol of nahco3 1 mol of hcl is required, so for 0.0053 mol of nahco3 we need 0.0053 mol.So total mol of hcl is,0.0053+0.0106= 0.0159 mol of hcl is needed...i.e. in .1M hcl .1mol is present in 1000ml of sol.Therefore 0.0159 is present in 159ml sol..