Answered By Leonardo F

First, we have to write the ionization chemical equation for the methanoic acid:

 $HCOOH\rightarrow H^++HCOO^-$HCOOH→H⁺+HCOO⁻ 

So, we know the proportion is 1:1:1. Now, with the pH, we can calculate the concentration of H+ ions:

 $pH=-log\left(\left[H^+\right]\right)$pH=−log([H⁺]) 

 $2.456=-log\left(\left[H^+\right]\right)$2.456=−log([H⁺]) 

Solving for the concentration of H+ ions, we have:

  $\left[H^+\right]\approx0.003499mol.L^{-1}$[H⁺]≈0.003499mol.L⁻¹ 

Given that the molar proportion of the methanoic acid to the H+ ions is 1:1, we know that the concentration of the methanoic acid must be also:

 $\left[HCOOH\right]=0.003499mol.L^{-1}$[HCOOH]=0.003499mol.L⁻¹ 

We know that the volume (V) used of methanoic acid is:

  $V=1.0\times10^{-5}mL=1.0\times10^{-8}L$V=1.0×10⁻⁵mL=1.0×10⁻⁸L 

Multiplying the volume used of methanoic acid and its concentration, we have the number of moles (n) of the acid required:

 $n=\left(0.003499mol.L^{-1}\right)\left(1.0\times10^{-8}L\right)=3.5\times10^{-11}mol$n=(0.003499mol.L⁻¹)(1.0×10⁻⁸L)=3.5×10⁻¹¹mol 

Given that the molar mass of the methanoic acid is 46.02538 g/mol, we can calculate the mass (m) necessary:

 $m=\left(3.5\times10^{-11}mol\right)\left(46.02538g.mol^{-1}\right)=1.6\times10^{-9}g$m=(3.5×10⁻¹¹mol)(46.02538g.mol⁻¹)=1.6×10⁻⁹g 

The mass is approximately 1.6 nanograms.