Answer
$[Tl^+] = 0.0181M$
$[OH^-] = 0.0181M$
$[H_3O^+] = 5.53 \times 10^{- 13}M$
Work Step by Step
1. Calculate the molar mass $(TlOH)$:
204.38* 1 + 16* 1 + 1.01* 1 = 221.39g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 2}{ 221.39}$
$n(moles) = 9.034\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
- 500ml = 0.500L
$ C(mol/L) = \frac{ 9.034\times 10^{- 3}}{ 0.500} $
$C(mol/L) = 0.01807$
- Since TlOH is a strong base:$[OH^-] = [Tl^+] = [TlOH] = 0.01807M$
4. Now, use the $K_w$ to find the hydronium concentration:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.807 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.807 \times 10^{- 2}}$
$[H_3O^+] = 5.534 \times 10^{- 13}M$
