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Your enamel is composed of a calcium hydroxyphosphate ($\ce{Ca5(PO4)3OH}$), also known as mineral apatite. This “apatite” is slowly broken down into calcium phosphate and water as illustrated in the chemical equation below: $$\ce{Ca5(PO4)3OH (s, tooth~enamel) + H+ (aq, acid) -> Ca5(PO4)3+ (aq) + H2O (l)}$$ In this experiment, you are trying to determine the acidity (or the hydrogen ion concentration) of some drinks, therefore, will be using a known concentration of a base to help you in this process. Recall that when acids and bases react, the neutral substance, water, is formed. An example of acetic acid and the base sodium hydroxide is given below: $$\ce{CH3COOH (aq) + NaOH (aq) -> CH3COONa (aq) + H2O (l)}$$

In this experiment you will use a sodium hydroxide solution to complete the titration process. If $\pu{100.0 mL}$ of a $\pu{0.20 M}$ $\ce{NaOH}$ solution is required, determine quantity of solute and solvent as well as a procedure to accomplish this goal of making a solution. Include details such as equipment and specific measurements that must occur. You are testing (lemonade, Gatorade, or ginger ale).

I got that I need $\pu{0.8 g}$ to make the solution.
Then I did the lab and got this data:

\begin{array}{lrrrr} \text{Drink} & \text{Vol. of Drink} & \text{Init. Vol. NaOH} & \text{Final Vol. NaOH} & \text{Vol.-Difference} \\\hline \text{Lemonade} & \pu{25 mL} & \pu{6.0 mL} & \pu{12.0 mL} & \pu{6.0 mL}\\ \text{Gatorade} & \pu{25 mL} & \pu{16.5 mL} & \pu{19.5 mL} & \pu{3.0 mL}\\ \text{Ginger Ale} & \pu{25 mL} & \pu{12.0 mL} & \pu{16.5 mL} & \pu{4.5 mL} \end{array}

With this information I need to answer these questions:

  1. Calculate and record the amount of $\ce{NaOH}$ used for each trial. Write the amount of $\ce{NaOH}$ used for each acid (lemonade, gatorade, and ginger ale).
  2. For each acid calculate the concentration $[\ce{M}]$ for each acid. Show all your work.
  3. Using the molarity of each acid calculated in #2, determine the pH of each solution.
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closed as off-topic by Martin - マーチン Mar 25 '16 at 21:00

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  1. With the volume of both solutions and the concentration of NaOH, you can calculate the unknown concentration with the following equation: $$M_{1}V_{1}=M_{2}V_{2}$$

    Consider the lemonade trial as an example.

    The concentration of your $\ce{NaOH}$ was $0.20~\mathrm{M}$ and in your log, you used a total of $6~\mathrm{mL}$ to reach the endpoint of your titration.
    $$\frac{0.20~\mathrm{M}_\ce{NaOH}\times 0.006~\mathrm{L}_\ce{NaOH}}{0.025~\mathrm{L}_\text{Lemomade}} = {0.048~\mathrm{M}_\text{Lemonade}}$$

    By dividing $V_{2}$, which is the volume of your analyte, you determine the unknown concentration. Be sure to convert $\mathrm{mL}$ to $\mathrm{L}$ in any calculations pertaining to molarity. Alternatively, you can convert $\mathrm{mol}$ to $\mathrm{mmol}$ for the same effect.

  2. To calculate the pH of a strong acid, this is easy: $$-\log [\ce{H+}] = \mathrm{pH}$$

    However, we are working with weak acids. Thus, we will need the $K_\mathrm{a}$ in order to determine the actual pH. $$K_\mathrm{a} =\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

    Continuing with the lemonade experiment, the accepted $K_\mathrm{a}$ of citric acid is $7.4\times{10}^{-4}$. Therefore: \begin{align} [\ce{H+}] &= \sqrt\frac{7.4\times10^{-4}}{0.048~\mathrm{M}_\text{Lem.}}\\ [\ce{H+}] &= 0.015~\mathrm{M} \end{align}

    And finally, we take the negative log of the dissociated hydronium to get the pH. $$-\log{[0.015~\mathrm{M}]}=1.8$$

    The pH of the lemonade was 1.8. The accepted equilibrium constants for each of the acids can be found online. The main acid in lemonade is citric acid, so search for the $K_\mathrm{a}$ value of citric acid.

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