Multiple Choice
What minimum volume of 0.248 M potassium iodide solution is required to completely precipitate all of the lead in 190.0 mL of a 0.100 M lead(II) nitrate solution?
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6. Chemical Quantities & Aqueous Reactions
Solution Stoichiometry
Multiple Choice
What volume (in mL) of 0.2000 M HCl is required to completely neutralize 50.00 mL of 0.8000 M NaOH?
A
400.0 mL
B
50.0 mL
C
100.0 mL
D
200.0 mL
Verified step by step guidance1
Write the balanced chemical equation for the neutralization reaction: \(\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}\).
Identify the molarity and volume of the base (NaOH): \(M_{\mathrm{NaOH}} = 0.8000\,M\), \(V_{\mathrm{NaOH}} = 50.00\,mL\).
Calculate the moles of NaOH using the formula: \(\text{moles} = M \times V\), where volume is in liters. So, \(\text{moles}_{\mathrm{NaOH}} = 0.8000 \times \frac{50.00}{1000}\).
Since the reaction is a 1:1 mole ratio between HCl and NaOH, the moles of HCl required are equal to the moles of NaOH calculated.
Calculate the volume of HCl needed using its molarity: \(V_{\mathrm{HCl}} = \frac{\text{moles}_{\mathrm{HCl}}}{M_{\mathrm{HCl}}} = \frac{\text{moles}_{\mathrm{NaOH}}}{0.2000}\). Convert this volume to mL by multiplying by 1000.
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