## General Chemistry (4th Edition)

Published by University Science Books

# Chapter 20 The Properties of Acids and Bases - Problems - Page 770: 3

#### 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: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1.807 \times 10^{- 2} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1.807 \times 10^{- 2}}$ $[H_3O^+] = 5.534 \times 10^{- 13}M$

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