A 30.0 cm3 sample of a gastric juice was neutralised using solid Mg(OH)2(s). The mass of the Mg(OH)2(s) was 0.341 g. What was the concentration of hydrochloric acid, in mol dm–3, in this sample of the gastric juice?State your answer to three significant figures and units are not required. M Mg(OH)2 = 58.32 g mol–1.Answer: mol dm–3
Question
A 30.0 cm3 sample of a gastric juice was neutralised using solid Mg(OH)2(s). The mass of the Mg(OH)2(s) was 0.341 g. What was the concentration of hydrochloric acid, in mol dm–3, in this sample of the gastric juice?State your answer to three significant figures and units are not required. M Mg(OH)2 = 58.32 g mol–1.Answer: mol dm–3
Solution
To solve this problem, we need to follow these steps:
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First, we need to calculate the number of moles of Mg(OH)2. We can do this using the formula: moles = mass / molar mass. In this case, the mass of Mg(OH)2 is 0.341 g and the molar mass is 58.32 g/mol. So, the number of moles of Mg(OH)2 is 0.341 g / 58.32 g/mol = 0.00585 mol.
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Next, we need to understand the reaction between Mg(OH)2 and HCl. The balanced chemical equation is: Mg(OH)2 + 2HCl -> MgCl2 + 2H2O. From this equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl. Therefore, the number of moles of HCl in the gastric juice is 2 * 0.00585 mol = 0.0117 mol.
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Finally, we need to calculate the concentration of HCl. The concentration is given by the formula: concentration = moles / volume. The volume of the gastric juice is 30.0 cm3, but we need to convert this to dm3 (1 dm3 = 1000 cm3), so the volume is 30.0 cm3 / 1000 = 0.03 dm3. Therefore, the concentration of HCl is 0.0117 mol / 0.03 dm3 = 0.39 mol/dm3.
So, the concentration of hydrochloric acid in the gastric juice is 0.39 mol/dm3.
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