Question

Do question "a" of the calculations part. Question "a" is in the image. The topic is about Titration.

Answer 1

Answer:

Trial 1; 0.102M

Trial 2: 0.094M

Trial 3: 10M

Explanations:

According to the dilution formula;

$C_1V_1=C_2V_2$

where:

• C1 and C2 are the initial and final concentration

,

• V1 and V2 are the initial and final volume

For the Trial 1:

• Concentration of HCl C1 = 0.1M

,

• Volume of HCl V1 = 10mL

,

• Volume of NaOH V2 = Vf - Vi = 10.3 - 0.5

,

• Volume of NaOH V2 = 9.8mL

Using the formula to determine molarity of base C2

$\begin{gathered} C_2=\frac{C_1V_1}{V_2} \\ C_2=\frac{0.1\times10}{9.8} \\ C_2=0.102M \end{gathered}$

For the Trial 2:

• Concentration of HCl C1 = 0.1M

,

• Volume of HCl V1 = 10mL

,

• Volume of NaOH V2 = Vf - Vi = 20.9 - 10.3

,

• Volume of NaOH V2 = 10.6mL

$\begin{gathered} C_2=\frac{0.1\times10}{10.6} \\ C_2=0.094M \end{gathered}$

For the Trial 3:

• Concentration of HCl C1 = 0.1M

,

• Volume of HCl V1 = 10mL

,

• Volume of NaOH V2 = Vf - Vi = 30.0 - 20.9

,

• Volume of NaOH V2 = 0.1mL

$\begin{gathered} C_2=\frac{0.1\times10}{0.1} \\ C_2=10M \end{gathered}$