Question

Baking soda (sodium hydrogen carbonate, ${\mathbf{NaHCO}}_{\mathbf{3}}$is used in baking because it reacts with acids in foods to form carbonic acid$\left(H_2{\mathrm{CO}}_3\right)$, which in turn decomposes to water and carbon dioxide. In a batter, the carbon dioxide appears as gas bubbles that cause the bread or cake to rise.

(a) A rule of thumb in cooking is that $\frac{\mathbf{1}}{\mathbf{2}}$teaspoon of baking soda is neutralized by 1 cup of sour milk. The acid component of sour milk is lactic acid${\mathbf{HC}}_3{\mathbf{H}}_5{\mathbf{O}}_3$. Write an equation for the neutralization reaction.

(b) If the density of baking soda is $\mathbf{2}\mathbf{.}\mathbf{16} \mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$. Calculate the concentration of lactic acid in the sour milk, in moles per liter. Take$\mathbf{1} \mathbf{cup}\mathbf{=}\mathbf{236}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{mL}\mathbf{=}\mathbf{48} \mathbf{tea} \mathbf{spoons}$.

(c) Calculate the volume of carbon dioxide that is produced at 1 atm pressure and${\mathbf{350}}^{\mathbf{o}}\mathbf{F}\left({\mathbf{177}}^{\mathbf{o}}\mathbf{C}\right)$from the reaction of localid="1663749359145" $\frac{\mathbf{1}}{\mathbf{2}}$ teaspoon of baking soda.

Short answer

1. The equation for the neutralization equation is,$$\begin{gathered} {\mathrm{NaHCO}}_3+{\mathrm{HC}}_3{\mathrm{H}}_5{\mathrm{O}}_3\to {\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{NaC}}_3{\mathrm{H}}_5{\mathrm{O}}_3 \\ {\mathrm{H}}_2{\mathrm{CO}}_3\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$
2. The concentration of lactic acid in the sour milk in moles per liter is $0.266\mathrm{M}$.
3. The volume of carbon dioxide is $2.27L$.

Step-by-step solution

Step 1:Subpart a)

The baking soda is neutralized by lactic acid to form carbonic acid and sodium lactate products.
The equation for the neutralization reaction is:

$$\begin{gathered} {\mathrm{NaHCO}}_3+{\mathrm{HC}}_3{\mathrm{H}}_5{\mathrm{O}}_3\to {\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{NaC}}_3{\mathrm{H}}_5{\mathrm{O}}_3 \\ {\mathrm{H}}_2{\mathrm{CO}}_3\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Subpart b)

We are asked to calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of$\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.$.
The rule of thumb in baking is that one cup of sour milk neutralizes 1/2 teaspoon of baking soda.
Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters).

$\mathrm{c}\left(\mathrm{molarity}\right)=\frac{\mathrm{n}\left(\mathrm{moles}\mathrm{of}\mathrm{solute}\right)}{\left(\mathrm{V}\right)}\left(\mathrm{L}\right)$

We're given the density of baking soda is $2.16\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{cm}}^3$}\right.$.

$\mathrm{density}=\frac{\mathrm{m}\left(\mathrm{mass}\right)}{\mathrm{V}\left(\mathrm{volume}\right)}$

Convert teaspoon (tsp) to milliliters (mL) and 1 cup to liters (L).

role="math" localid="1663750057350" $$\begin{gathered} 1\mathrm{tsp}=4.92\mathrm{mL}=4.92{\mathrm{cm}}^3 \\ 1 \mathrm{cup}=236.6\mathrm{mL}=48 \mathrm{tea} \mathrm{spoons} \\ 1\mathrm{mL}={10}^{-3}\mathrm{L} \end{gathered}$$

We are given tsp of sodium bicarbonate, ${\mathrm{NaHCO}}_3$.

$$\begin{gathered} \mathrm{V}\left({\mathrm{NaHCO}}_3\right)=0.5\mathrm{tsp}\times \left(\frac{4.92{\mathrm{cm}}^3}{1\mathrm{tsp}}\right) \\ \mathrm{V}\left({\mathrm{NaHCO}}_3\right)=2.46{\mathrm{cm}}^3 \end{gathered}$$

We are given 1 cup of sour milk.

$$\begin{gathered} \mathrm{V}\left(\mathrm{sour}\mathrm{milk}\right)=1\mathrm{cup}\times \left(\frac{236\mathrm{mL}}{1\mathrm{cup}}\right)\left(\frac{{10}^{-3}\mathrm{L}}{1\mathrm{mL}}\right) \\ \mathrm{V}\left(\mathrm{sour}\mathrm{milk}\right)=0.2366\mathrm{L} \end{gathered}$$

Calculate the moles of baking soda (sodium bicarbonate, role="math" localid="1663751778450" ${\mathrm{NaHCO}}_3$). To determine the moles of${\mathrm{NaHCO}}_3$, we need their molar mass.
The molar mass of${\mathrm{NaHCO}}_3$ is$84\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$.

Calculate the moles of ${\mathrm{NaHCO}}_3$:

$\mathrm{n}\left({\mathrm{NaHCO}}_3\right)=2.46{\mathrm{cm}}^3\times \left(\frac{2.16{\mathrm{gNaHCO}}_3}{{\mathrm{cm}}^3{\mathrm{NaHCO}}_3}\right)\left(\frac{1{\mathrm{molNaHCO}}_3}{84{\mathrm{gNaHCO}}_3}\right)$

Divide those numbers, and we get:

$\mathrm{n}\left({\mathrm{NaHCO}}_3\right)=0.063\mathrm{mo}l$

Perform a mole-to-mole comparison between lactic acid and sodium carbonate (1 mol lactic acid: 1 mol sodium carbonate)

$\mathrm{n}\left(\mathrm{moles}\mathrm{of}\mathrm{lactic}\mathrm{acid}\right)=\mathrm{n}\left({\mathrm{NaHCO}}_3\right)=0.063\mathrm{mol}$

Calculate the molarity:

$\mathrm{c}=\frac{\mathrm{n}}{\mathrm{V}}$

Plug the data in:

$\mathrm{c}=\frac{0.063\mathrm{mol}}{0.2366\mathrm{L}}$

Divide the numbers, and we get the final concentration:

$\mathrm{c}=0.266\mathrm{M}$

Subpart c)

n(number of moles for lactic acid) = 0.063 mol
Looking at the neutralization equation, we see that there is the same number of moles for lactic acid and carbon dioxide, therefore:

$\mathrm{n}\left({\mathrm{CO}}_2\right)=0.063\mathrm{mol}$

$Temperature=350°\mathrm{F}\left(177°\mathrm{C}\right)=449.82\mathrm{K}$

To calculate the volume, use the formula:

$\mathrm{pV}=\mathrm{nRT}$

Plug the data in:

$$\begin{gathered} 1\mathrm{atm}\times \mathrm{V}=0.063\mathrm{mol}\times 0.0802\frac{\mathrm{L}/\mathrm{atm}}{\mathrm{Kmol}}\times 449.82\mathrm{K} \\ \mathrm{V}\left({\mathrm{CO}}_2\right)=2.27\mathrm{L} \end{gathered}$$