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aleksandr82 [10.1K]
2 years ago
7

Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a

dded to a sample of Mg metal. The reaction goes to completion and all of the Mg metal is gone. 27.08 mL of 0.30 M NaOH solution is required to reach the end point when titrating the remaining HCl.
Chemistry
1 answer:
musickatia [10]2 years ago
3 0

Answer:

Approximately 1.876 \times 10^{-3}\; \rm mol.

Explanation:

Convert both volumes to standard units (that is: liters.)

  • 10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L.
  • 27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L.

Number of moles of \rm HCl initially present (in the 10.00\; \rm mL solution at 1.00\; \rm M.)

n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol.

Number of moles of \rm NaOH from the titration:

n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol.

\rm NaOH neutralizes \rm HCl at a 1:1 ratio:

\rm HCl + NaOH \to NaCl + H_2O.

Hence, n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol.

\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}.

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