Question

9. A 50 ml solution of unknown base contains 0.0012 mols of base (2.94 g/L) in solution.
Calculate the molar mass of the solution, and use that to calculate the concentration and
pH of the solution.

Answer 1

Solution.

$V=50mL=0.05L;$

$\nu_x=0.0012mol;$

$\rho_x=2.94g/L;$

$\nu_x=\dfrac{V_x}{V_m};\implies V_x=\nu_xV_m;$ $V_m$ - volume of one mole;

$V_x=0.0012mol\sdot22.4L/mol=0.02688L;$

$\rho_x=\dfrac{m_x}{V_x};\implies m_x=\rho_xV_x;$

$m_x=2.94g/L\sdot0.02688L=0.079g;$

$\nu_x=\dfrac{m_x}{M_x} \implies M_x=\dfrac{m_x}{\nu_x}$ ;

$M_x=\frac{0.079g}{0.0012mol}=65.856g/mol;$

$C_x=\dfrac{\nu_x}{V};$ $C_x=\dfrac{0.0012mol}{0.05L}=0.024mol/L;$

$pH=-lgC_x; pH=-lg0.024=lg41.67=1.62;$

Answer:$M_x=65.856g/mol;$

$C_x=0.024mol/L;$

$pH=1.62;$