Calcium carbonate is often used as an antacid. Your stomach acid is composed of HCl at a pH of 1.5. If you ate toooo much Turkey and need to

Question

Calcium carbonate is often used as an antacid. Your stomach acid is composed of HCl at a pH of 1.5. If you ate toooo much Turkey and need to neutralize 15.0 mL of stomach acid, how many grams of calcium carbonate would you need to take

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Phúc Điền 5 years 2021-07-26T15:31:52+00:00 1 Answers 27 views 0

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  1. Amity
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    2021-07-26T15:33:21+00:00
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    Answer: 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

    Explanation:

    pH is defined as the negative logarithm of hydrogen ion concentration present in the solution

    pH=-\log [H^+]      …..(1)

    Given value of pH = 1.5

    Putting values in equation 1:

    1.5=-\log[H^+]

    [H^+]=10^{(-1.5)}=0.0316M

    Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

    \text{Molarity of solution}=\frac{\text{Number of moles of solute}\times 1000}{\text{Volume of solution (mL)}}       …..(2)

    We are given:

    Volume of solution = 15.0 mL

    Molarity of HCl = 0.0316 M

    Putting values in equation 2:

    0.0316=\frac{\text{Moles of HCl}\times 1000}{15.0}\\\\\text{Moles of HCl}=\frac{0.0316\times 15.0}{1000}=4.74\times 10^{-4}mol

    The chemical equation for the reaction of HCl and calcium carbonate follows:

    2HCl+CaCO_3\rightarrow H_2CO_3+CaCl_2

    By the stoichiometry of the reaction:

    2 moles of HCl reacts with 1 mole of calcium carbonate

    So, 4.74\times 10^{-4}mol of HCl will react with = \frac{1}{2}\times 4.74\times 10^{-4}=2.37\times 10^{-4}mol of calcium carbonate

    The number of moles is defined as the ratio of the mass of a substance to its molar mass.

    \text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

    Moles of calcium carbonate = 2.37\times 10^{-4}mol

    Molar mass of calcium carbonate = 100.01 g/mol

    Putting values in the above equation:

    \text{Mass of }CaCO_3=(2.37\times 10^{-4}mol)\times 100.01g/mol\\\\\text{Mass of }CaCO_3=0.0237g

    Hence, 0.0237 g of calcium carbonate would be required to neutralize the given amount of HCl

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