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Given Ka of HCOOH = 1.8*10^-4

We know that,pKa = -log Ka

=-log 1.8*10^-4

=4*log 1.8

=4*0.25

pKa=1

ByHenderson Hasselbach  equation,pH = pKa + log([A-]/[HA])

=pH of HCOOH = pKa of HCOOH + log([salt]/[acid])

=pH of HCOOH = 1+log([10^-2]/[10^-2])

therefore,,,pH of HCOOH =1

Concentration of HCOOH = 10^-pH

=10^-1

Hope it helps!!!!