Question

What is the pH of a 25°C solution containing 10-3 moles NaHCO3 and 2×10-3 moles of NH3 per liter of seawater? Assume that the dissolved solids concentration for seawater is 35,000 mg/L. Assume closed system.

Answer 1

Here, 0.001 M NaHCO₃ and 0.002 M NH₃ both are together in a litre solution.

Therefore, the total concentration of the solution is,

0.001 + 0.002 = 0.003 M per litre solution

= 3 $\times$ 10^-3 M per litre solution

Both components used to prepare the solution are base, therefore, pOH first calculated for the total concentration.

pOH = -log [OH^-]

= -log [3 $\times$ 10^-3]

= -[-3+ 0.4771]

= -[- 2.5229]

= 2.5229

The equation for the calculation of pH is

pH + pOH = 14,

the value of pOH calcuated is substituted into the above equation,

Thus, pH + 2.5229 = 14

pH = 14 - 2.5229

pH = 11.4771

Therefore, the pH of the solution containing 1 $\times$ 10^-3 M NaHCO₃ and 2 $\times$ 10^-3 M NH₃ is 11.4771