## Chemistry 10th Edition

(a) $[K^+] = [OH^-] = 0.01485M$ (b) $[Ba^{2+}] = 5.849 \times 10^{-3}M$ $[OH^-] = 0.0117M$ (c) $[Ca^{2+}] = 0.07678M$ $[NO_3^{-}] = 0.1536M$
(a) 1. Calculate the molar mass: Molar Mass ($KOH$): 39.1* 1 + 16* 1 + 1.01* 1 = 56.11g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.25}{ 56.11}$ $n(moles) = 0.02228$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.02228}{ 1.5}$ $C(mol/L) = 0.01485$ $KOH$ is as strong base, so $[K^+] = [OH^-] = [KOH] = 0.01485M$ (b) 1. Calculate the molar mass: Molar Mass ($Ba(OH)_2$): 137.3* 1 + 16* 2 + 1.01* 2 = 171.32g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.2505}{ 171.32}$ $n(moles) = 1.462\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1.462\times 10^{- 3}}{ 0.25}$ $C(mol/L) = 5.849\times 10^{- 3}M$ $Ba(OH)_2$ is a strong base, therefore: $[Ba^{2+}] = [Ba(OH)_2] = 5.849 \times 10^{-3}M$ $[OH^-] = [Ba(OH)_2] * 2= 5.849 \times 10^{-3} * 2 = 0.0117M$ (c) 1. Calculate the molar mass: Molar Mass ($Ca(NO_3)_2$): 40.08* 1 + 14.01* 2 + 16* 6 = 164.1g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.26}{ 164.1}$ $n(moles) = 7.678\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 7.678\times 10^{- 3}}{ 0.1}$ $C(mol/L) = 0.07678M$ $Ca(NO_3)_2$ is a salt, therefore: $[Ca^{2+}] = 0.07678M$, and $[NO_3^{-}] = 2 * 0.07678 = 0.1536M$