Question

how many grams of H3BO3 (s) will be required for the complete reaction of 1.55 ml of 28.6 M HF (aq)?

Answer 1

$H_3BO_3 + HF = H[BF_4] + H_2O$

As per reaction ,

$M_2V_2=$ Moles of $HF=$ Moles of $H_3BO_3=28.6\times 1.55\ milimoles=$ $44.33\times 10^{-3} \ moles$

Weight of $H_3BO_3= Moles\times Molar \ mass=$

$44.33 \times 61.83 \times 10^{-3}=$ $2.741 \ gm$