Answer:
Explanation:
Here, we want to answer some questions about titration
We shall be answering questions for sample 1 as follows
a) Mass of KHP in sample 1
That would be the difference between mass of beaker + KHP after first sample taken minus mass of beaker + KHP before sample taken
Mathematically, we have that as:
[tex]4.5155\text{ - 4.0032 = 0.5123 g}[/tex]
b) Moles of KHP
Mathematically, we have that as:
[tex]\begin{gathered} \frac{mass\text{ of KHP}}{molar\text{ mass of KHP}} \\ \\ molar\text{ mass of KHP is 204.22 g/mol} \\ \\ Number\text{ of moles:} \\ \frac{0.5123}{204.22}\text{ = 0.0025 mole} \end{gathered}[/tex]
c) Volume of NaOH utilized
We can get that from the given table
We have the value as 22.75 mL/1000 = 0.2275 L
d) Molarity of NaOH from sample 1
Let us write the equation of reaction;
[tex]KHP\text{ + NaOH }\rightarrow\text{ KNAP + H}_2O[/tex]
We have the general equation to use as:
[tex]\frac{C_aV_a}{n_a}\text{ = }\frac{C_bV_b}{n_b}[/tex]
But we need to calculate the concentration of the acid
We have the volume of acid used as 22.86 - 0.11 = 22.75 mL = 0.2275 L
The molarity would be:
[tex]\frac{mass\text{ of KHP}}{molar\text{ mass of KHP}}\text{ }\times\text{ }\frac{1}{vol\text{ in L}}[/tex][tex]\frac{0.0025}{0.2275}\text{ = 0.011 M}[/tex]
From the dilution equation above:
Ca is KHP molarity = 0.011 M
Va is KHP volume = 0.2275 L
Cb is NaOH molarity = ?
Vb is NaOH volume = 0.2275
na is the number of moles of acid in the balanced equation which is 1
nb is the number of moles of base is balanced equation which is 2
Substituting the values:
[tex]\begin{gathered} \frac{0.011\text{ }\times\text{ 0.2275}}{1}\text{ = }\frac{C_b\times0.2275}{1} \\ \\ C_b\text{ = 0.011 M} \end{gathered}[/tex]
