Using the following chemical equation and your vast knowledge of stoichiometry, calculate the exact molarity of the NaOH. Hint—it should be around 0.1 M.KHP(aq) + NaOH(aq) NaKP(aq) + H2O(liquid)a)Convert the grams of KHP into moles of KHP using the formula weight of 204.2 grams per mole.b)Convert moles of KHP into moles of NaOH using the stoichiometry given in the above equation.c)Calculate the molar concentration of the NaOH by dividing the number of moles of NaOH calculated immediately above by the volume of NaOH used (measured using the buret). Be sure to convert the volume from mL to L.d)Average the three values obtained to determine the average molar the concentration of the NaOH. GET INFORMATION FROM THE PICTURE

[tex]\begin{gathered} \text{Average mass = }\frac{mass\text{ 1 + mass 2 + mass 3}}{3} \\ \text{Average mass = }\frac{0.55\text{ + 0.56 + 0.56}}{3} \\ \text{Average mass = }\frac{1.166}{3} \\ \text{Average mass = 0.3886 grams} \end{gathered}[/tex]

The average mass of KHP used is 0.3886grams

[tex]\begin{gathered} \text{Mole = }\frac{\text{ reacting mass}}{\text{molar mass}} \\ \text{reacting mass = 0}.3886\text{ grams} \\ \text{Molar mass = 2}04.2\text{ grams/mol} \\ \text{Mole = }\frac{0.3866\text{ }}{204.2} \\ \text{Mole of KHP = 0.0019 mole} \end{gathered}[/tex]

The mole of KHP is 0.0019 mole

PART B

According to the balanced equation, the stoichiometry ratio of KHP to NaOH is 1: 1

Let the mole of NaOH be x

[tex]\begin{gathered} 1\text{ : 1 = }0.0019\text{ : x} \\ \frac{1}{1}\text{ = }\frac{0.0019}{x} \\ \text{Cross multiply} \\ 1\cdot\text{ x = 1 }\cdot\text{ 0.0019} \\ x\text{ = 0.0019 mole} \end{gathered}[/tex]

Hence, the mole of NaOH is 0.0019 mole

PART C

Given the following parameters

0. The volume of NaOH used in flask 1 = 26.70mL

,

1. The volume of NaOH used in flask 2= 27.09mL

,

2. The volume of NaOH used in flask 3 = 26.96mL

The next step is to convert the mL to L

[tex]1mL\text{ = 0.001L}[/tex]

For flask 1

[tex]\begin{gathered} 1mL\text{ = 0.001L} \\ \text{Let x be the volume of NaOH in L} \\ \text{ 1mL = 0.001L} \\ 26.70mL\text{ = xL} \\ \text{Cross multiply} \\ xL\cdot\text{ 1ml = 26.70mL }\cdot\text{ 0.001L} \\ x\text{ = }\frac{26.70\cdot\text{ 0.001}}{1} \\ x\text{ = 0.0267L} \end{gathered}[/tex]

Using the same conversion process

The volume of NaOH in L in flask 2 = 0.02709L

The volume of NaOH in L in flask 3 = 0.02696L

Hence, the molar concentration of the solution in each flask can be calculated as follows

[tex]\text{Molar concentration = }\frac{concentration}{\text{Volume}}[/tex]

For flask 1

Mole of NaOH = 0.0019 mole

Volume of NaOH = 0.0267L

[tex]\begin{gathered} \text{Molar concentration = }\frac{0.0019}{0.0267} \\ \text{Molar concentration = }0.0711\text{ mol/L} \end{gathered}[/tex]

For flask 2

Mole = 0.0019 mole

Volume = 0.02709L

[tex]\begin{gathered} \text{Molar concentration = }\frac{Concentration}{\text{volume}} \\ \text{Molar concentration = }\frac{0.0019}{0.02709} \\ \text{Molar concentration = 0.0701 mol/L} \end{gathered}[/tex]

For flask 3

Mole = 0.019mole

Volume = 0.02696 L

[tex]\begin{gathered} \text{Molar concentration = }\frac{concentration}{\text{volume}} \\ \text{Molar concentration = }\frac{0.0019}{0.02696} \\ \text{Molar concentration = 0.0704 mol/L} \end{gathered}[/tex]

PART D

Average molar concentration can be found using the below formula

[tex]\begin{gathered} \text{Average molar concentration = }\frac{0.0711\text{ + 0.0701 + 0.0704}}{3} \\ \text{Average molar concentration =}\frac{0.2116}{3} \\ \text{Average molar concentraion = 0.0705 mol/L} \\ \text{The average molar concentration of NaOH is 0.0705 mol/L} \end{gathered}[/tex]