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Multiple Choice
A student titrates 25.00 mL of HCl solution with 0.100 M NaOH. If it takes 32.50 mL of NaOH to reach the endpoint, how many moles of HCl were present in the original 25.00 mL of acid?
A
0.00325 mol
B
0.00250 mol
C
0.00350 mol
D
0.00300 mol
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1
Write the balanced chemical equation for the titration: \(\mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O}\). This shows a 1:1 mole ratio between HCl and NaOH.
Calculate the moles of NaOH used in the titration using the formula: \(\text{moles NaOH} = M_{\mathrm{NaOH}} \times V_{\mathrm{NaOH}}\), where \(M_{\mathrm{NaOH}}\) is the molarity of NaOH and \(V_{\mathrm{NaOH}}\) is the volume in liters.
Convert the volume of NaOH from milliliters to liters by dividing by 1000: \(V_{\mathrm{NaOH}} (L) = \frac{32.50\, \mathrm{mL}}{1000}\).
Since the mole ratio of HCl to NaOH is 1:1, the moles of HCl in the original solution are equal to the moles of NaOH used at the endpoint.
Report the moles of HCl calculated as the amount present in the original 25.00 mL of acid solution.