Question

Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate.

Short answer

The sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 mL of 1.0 M sodium bicarbonate is .

Step-by-step solution

Definition

The molar mass of a material is defined as the mass of the substance contained inone moleof the substance, and it is equal to the total mass of the component atoms. It may be stated mathematically as follows:

$Molarmass=\frac{Weighting}{numberofmoles}$

Determination of moles of sodium ions in the solution

Let’s calculate the moles of $N{a}_{2}C{O}_3$,

$$\begin{gathered} n\left(N{a}_{2}C{O}_3\right)=c\left(N{a}_{2}C{O}_3\right)\times V\left(N{a}_{2}C{O}_3\right) \\ =3.0M\times 0.07L \\ =0.21mol \end{gathered}$$

Let’s calculate the moles of sodium ions,

$$\begin{gathered} n_1\left(Na\right)=2\times \left(0.21mol\right) \\ =0.42mol \end{gathered}$$

Let’s calculatethe moles of $NaHC{O}_3$,

$n\left(NaHC{O}_3\right)=c\left(NaHC{O}_3\right)\times V\left(NaHC{O}_3\right)$

$$\begin{gathered} =1.0M\times 0.03L \\ =0.03mol \end{gathered}$$

Determination of moles of sodium ions

Let’scompute the moles of sodium ion,

$$\begin{gathered} n_2\left(Na\right)=1\times \left(0.03mol\right) \\ =0.03mol \\ n\left(Na\right)=0.42mol+0.03mol \\ =0.45mol \end{gathered}$$

So, the sodium ions concentration is given,

$$\begin{gathered} c\left(N{a}^{+}\right)=\frac{n\left(N{a}^{+}\right)}{V} \\ =\frac{n\left(N{a}^{+}\right)}{V_1+V_2} \\ =\frac{0.45mol}{0.1L} \\ =4.5M \end{gathered}$$

Therefore the concentration of sodium ionsis$4.5M$ .