Q19.135CP - Chemistry: Molecular Nature Of Matter And Change

Question

Environmental engineers use alkalinity as a measure of the capacity of carbonate buffering systems in water samples:

Alkalinity  $\mathbf{\text{(mol/L)=}}\mathbf{\left[}{\mathbf{\text{HCO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{-}}}\mathbf{\right]}\mathbf{\text{+2}}\mathbf{\left[}{{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{2-}}}\mathbf{\right]}\mathbf{\text{+}}\mathbf{\left[}{\mathbf{\text{OH}}}^{\mathbf{\text{-}}}\mathbf{\right]}\mathbf{\text{-}}\mathbf{\left[}{\mathbf{\text{H}}}^{\mathbf{\text{+}}}\mathbf{\right]}$

Find the alkalinity of a water sample that has a   $\mathbf{\text{pH}}$of  , ${{\mathbf{\text{26.0 mg/L CO}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{2-}}}$$\mathbf{\text{9.5,}}$and.${{\mathbf{\text{65.0 mg/L HCO}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$

Step-by-Step Solution

Answer

The alkalinity of a water sample is .${\text{ Alkalinity =1.2×10}}^{\text{-3}}\text{M}$

Step 1: Definition of Concept

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

 

Step 2:Find the alkalinity of a water sample

Considering the given information:

 $\text{γ}\left({\text{CO}}_{\text{3}}^{\text{2-}}\right)\text{=26mg/L=0.026 g/Lγ}\left({\text{HCO}}_{\text{3}}^{\text{-}}\right)\text{=65mg/L=0.065 g/LpH=9.5}$

We can use this dividing equation to find the molarity of .${\text{CO}}_{\text{3}}^{\text{2-}}$

 $\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]\text{=0.026}\frac{\text{g/100g}}{\text{ mol/L}}\text{=0.0026M}$

The molarity of   ${\text{HCO}}_{\text{3}}^{\text{-}}$can be calculated in the same way:

$\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]\text{=0.065}\frac{\text{g/100 g}}{\text{ mol/L}}\text{=0.0065M}$ 

We can calculate using the given $\text{pH}$  value.

 $\left[{\text{H}}^{\text{+}}\right]$:

$\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{\text{-pH}}{\text{=3.16×10}}^{\text{-10}}\text{M}$ 

We can now calculate :$\left[{\text{OH}}^{\text{-}}\right]$

 $\begin{array}{c}\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{\text{Kw}}{\left[{\text{H}}^{\text{+}}\right]}\\ {\text{=3.16×10}}^{\text{-5}}\text{M}\end{array}$

$\begin{array}{l}\text{ Alkalinity =}\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]\text{+2}\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]\text{+}\left[{\text{OH}}^{\text{-}}\right]\text{-}\left[{\text{H}}^{\text{+}}\right]\\ {\text{ Alkalinity =1.2×10}}^{\text{-3}}\text{M}\end{array}$

 

Therefore, the required value is .