Post-Lab Question (Show All Calculations) A student sets out to standardize a so
ID: 1034318 • Letter: P
Question
Post-Lab Question (Show All Calculations) A student sets out to standardize a solution of NaOH as was done in this experiment. The students dissolves 0.2442 g of KHP (204.2 g/mol) in DI water, adds 2 drops of indicator, and adds the Na0H solution drop wise till the equivalence point is reached. The equivalence point is reached when 15.6 mL of the NaOH solution is added. The neutralization reaction is: 1. KHP(aq) NaOH(aq) KNaP(aa) H200) How many moles of KHP are used in this experiment? Show your calculations here for full credit a. b. How many moles of NaOH were needed to neutralize this amount of KP? What is the molarity (mol/L) of the NaOH solution? Show your calculations here for full credit. c. 2. Using the standardized NaOH solution from question 1, the student titrates exactly 10.0 mL of an unknown monoprotic weak acid as was done in this experiment and obtains the following: Volume at equivalence point: Volume at one half equivalence point: 6.3 mL pH at the one half equivalence point: 4.72 a. What is the pKa of the weak acid used in this experiment? 12.6 mL What is the Ka of the weak acid used in this experiment? Show your calculations here for full credit. b. c. Based on the pKa value, what could the weak acid be?Explanation / Answer
1) a) mol of KHP=mass of KHP/molar mass of KHP=0.2442g/(204.2g/mol)=0.00119 mol
b)mol KHP/mol NaOH=1:1 mol ratio
So mol of NaOH required for full neutralization=0.00119 mol
c)Volume of NaOH needed to reach equivalence point=15.6ml=0.0156L [1L=1000ml]
Molarity of NaOH=mol NaOH/volume NaOH=0.00119mol/0.0156L=0.0763 M
Molarity of NaOH=0.0763 M