Question

1. based on your answers to questions 1 through 4 on the diagram and your knowledge of chemistry, in a laboratory experiment, a student uses titration to neutralize a 15.0 ml solution of 2.0 m naoh. the accepted value of 1.0 m hcl needed to neutralize the naoh solution is 30.0 ml. the data table below shows the three trials a student performed during the experiment and the recorded volume of hcl used each time. construct a mathematical representation and calculate the percent error for the volume of hcl.

trial 1: 29.0 ml
trial 2: 30.5 ml
trial 3: 31.2 ml

1. write and support a claim, citing quantitative evidences, to demonstrate which trial allowed for the smallest percent error.

claim:
evidence:
reasoning:

1. a student determines the density of an unknown metal by measuring its mass and volume. the student records the mass as 52.4 g and the volume as 6.91 ml.

a. show a mathematical representation and the calculated result of the density value.
b. the accepted density value of the metal is 7.87 g/ml. show a mathematical representation and the calculated result for the percent error.

Explanation:

Step1: Calculate the molar - ratio in the titration reaction

The reaction between HCl and NaOH is \(HCl + NaOH=NaCl + H_2O\), and the molar - ratio of \(HCl\) to \(NaOH\) is \(1:1\). The formula for the number of moles \(n = M\times V\) (where \(M\) is molarity and \(V\) is volume in liters). For \(NaOH\), \(M_{NaOH}=2.0\ M\) and \(V_{NaOH}=15.0\ mL = 0.015\ L\), so \(n_{NaOH}=M_{NaOH}\times V_{NaOH}=2.0\ mol/L\times0.015\ L = 0.03\ mol\). Since the molar - ratio of \(HCl\) to \(NaOH\) is \(1:1\), \(n_{HCl}=n_{NaOH}\).

Step2: Calculate the theoretical volume of \(HCl\) for each trial

For \(HCl\), \(M_{HCl}=1.0\ M\), and using \(V=\frac{n}{M}\), since \(n_{HCl}=0.03\ mol\), the theoretical volume \(V_{HCl}=\frac{0.03\ mol}{1.0\ mol/L}=0.03\ L = 30.0\ mL\).

Step3: Calculate the percent error for each trial

The percent - error formula is \(\text{Percent Error}=\frac{\vert\text{Experimental Value}-\text{Theoretical Value}\vert}{\text{Theoretical Value}}\times100\%\).

Trial 1:

\(V_{exp1}=29.0\ mL\), \(\text{Percent Error}_1=\frac{\vert29.0 - 30.0\vert}{30.0}\times100\%=\frac{1.0}{30.0}\times100\%\approx3.33\%\)

Trial 2:

\(V_{exp2}=30.5\ mL\), \(\text{Percent Error}_2=\frac{\vert30.5 - 30.0\vert}{30.0}\times100\%=\frac{0.5}{30.0}\times100\%\approx1.67\%\)

Trial 3:

\(V_{exp3}=31.2\ mL\), \(\text{Percent Error}_3=\frac{\vert31.2 - 30.0\vert}{30.0}\times100\%=\frac{1.2}{30.0}\times100\% = 4.0\%\)
The smallest percent error is for Trial 2.

For the metal density problem:

Step1: Calculate the experimental density

The formula for density is \(
ho=\frac{m}{V}\), where \(m = 52.4\ g\) and \(V = 6.91\ mL\), so \(
ho_{exp}=\frac{52.4\ g}{6.91\ mL}\approx7.58\ g/mL\)

Step2: Calculate the percent error

The accepted density \(
ho_{acc}=7.87\ g/mL\). The percent - error formula is \(\text{Percent Error}=\frac{\vert
ho_{exp}-
ho_{acc}\vert}{
ho_{acc}}\times100\%\)
\(\text{Percent Error}=\frac{\vert7.58 - 7.87\vert}{7.87}\times100\%=\frac{0.29}{7.87}\times100\%\approx3.69\%\)

Answer:

For the HCl - NaOH titration, the percent errors for Trial 1, Trial 2, and Trial 3 are approximately \(3.33\%\), \(1.67\%\), and \(4.0\%\) respectively. The trial with the smallest percent error is Trial 2.
For the metal density, the experimental density is approximately \(7.58\ g/mL\) and the percent error is approximately \(3.69\%\)