Question

Form calculations based on the photos. A. Mass of KHP in Sample 1,2,3,4B. Moles of KHP in Sample 1,2,3,4C. Volume of NaOH utilized in liters for Sample 1,2,3,4D. Molarity of NaOH from Sample 1,2,3,4

Answer 1

Step-by-step 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:

`4.5155\text{ - 4.0032 = 0.5123 g}`

b) Moles of KHP

Mathematically, we have that as:

`\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:} \\ (0.5123)/(204.22)\text{ = 0.0025 mole} \end{gathered}`

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;

`KHP\text{ + NaOH }\rightarrow\text{ KNAP + H}_2O`

We have the general equation to use as:

`(C_aV_a)/(n_a)\text{ = }(C_bV_b)/(n_b)`

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:

`\frac{mass\text{ of KHP}}{molar\text{ mass of KHP}}\text{ }*\text{ }\frac{1}{vol\text{ in L}}`
`(0.0025)/(0.2275)\text{ = 0.011 M}`

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:

`\begin{gathered} \frac{0.011\text{ }*\text{ 0.2275}}{1}\text{ = }(C_b*0.2275)/(1) \\ \\ C_b\text{ = 0.011 M} \end{gathered}`