Answer
$0.5686$g of $HI.$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.75}$
$[H_3O^+] = 1.778 \times 10^{- 2}$
2. Since HI is a strong acid, $[HI] = [H_3O^+] =1.778 \times 10^{-2}M$.
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 0.01778 * 0.25$
$n(moles) = 4.446\times 10^{- 3}$
4. Find the mass value in grams:
1.01* 1 + 126.9* 1 = 127.91g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 127.91 * 4.446\times 10^{- 3}$
$mass(g) = 0.5686$
