Answer
(a) $[K^+] = [OH^-] = 0.01485M$
(b) $[Ba^{2+}] = 5.849 \times 10^{-3}M$
$[OH^-] = 0.0117M$
(c) $[Ca^{2+}] = 0.07678M$
$[NO_3^{-}] = 0.1536M$
Work Step by Step
(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 $[NO_3^{-}] = 2 * 0.07678 = 0.1536M$
