Answer
1.798g of $HI$.
Work Step by Step
1. Calculate the hydronium concentration.
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.25}$
$[H_3O^+] = 5.623 \times 10^{- 2}M$
2. Since [HI] is a strong acid, $[HI] = [H_3O^+] = 5.623 \times 10^{- 2}M$.
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 0.05623 * 0.25$
$n(moles) = 0.01406$
4. Find the mass value in grams:
1.01* 1 + 126.9* 1 = 127.91g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 127.91 * 0.01406$
$mass(g) = 1.798$
