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What volume in milliliters (mL) of an HCl solution with a pH of 1.84 can be neutralized by 34.0 mg of CaCO3?

User Mgrandi
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Answer:

49.0 mL

Step-by-step explanation:

The balanced chemical equation for the reaction between HCl and CaCO3 is:

HCl + CaCO3 → CaCl2 + CO2 + H2O

From the equation, we can see that 1 mol of CaCO3 reacts with 2 moles of HCl. We can use the molar mass of CaCO3 (100.09 g/mol) to find the number of moles of CaCO3 in 34.0 mg:

34.0 mg CaCO3 × (1 g / 1000 mg) × (1 mol / 100.09 g) = 3.398 × 10^-4 mol CaCO3

Since 1 mol of CaCO3 reacts with 2 moles of HCl, we need twice the number of moles of HCl to neutralize all of the CaCO3:

2 × 3.398 × 10^-4 mol HCl = 6.796 × 10^-4 mol HCl

The pH of the HCl solution tells us that the concentration of H+ ions is:

[H+] = 10^-pH

[H+] = 10^-1.84

[H+] = 1.39 × 10^-2 M

The number of moles of HCl needed to neutralize the CaCO3 can be calculated using the volume of the HCl solution and its concentration:

Molarity (M) = moles (mol) / volume (L)

volume (L) = moles (mol) / Molarity (M)

volume (L) = 6.796 × 10^-4 mol / 1.39 × 10^-2 M

volume (L) = 0.049 L

Finally, we can convert the volume to milliliters:

volume (mL) = 0.049 L × (1000 mL / 1 L)

volume (mL) = 49.0 mL

Therefore, 49.0 mL of the HCl solution with a pH of 1.84 can be neutralized by 34.0 mg of CaCO3.

User Michalis
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