a) How many moles of KHP did you you weigh in Trial 1?
Be sure to show all work with proper the proper number of significant figures and units. Show numbers with short descriptors. For example, 0.2000 g (mass KHP)/204.22 g/mol (MW KHP) = 0.0009793 mol KHP
b) How many moles of HCl did you use in trial 1? If the back titration was not necessary indicate this in your response.
Show all work with the proper number of significant figures and units. Show numbers with short descriptors. For example 1.00 mL (volume HCl) x (1 L/1000 mL) x 0.100 M (molarity HCl) = 0.0001 mol HCl
c) What was the total moles of H+ in the sample including KHP and HCl? Show all work with the proper number of significant figures and units. Show numbers with short descriptors.
a) Moles of KHP in Trial 1= 0.2520 g (mass KHP) x (1 mol KHP / 204.22 g/mol (MW KHP)) = 0.001234 mol KHP
b) Moles of HCl used in Trial 1= 0.60 mL (volume HCl) x (1 L / 1000 mL) x 0.2517 M (molarity HCl) = 0.00015 mol HCl
c) Total moles of H+ in the sample including KHP and HCl= 0.001234 mol H+ (molar KHP) + 0.00015 mol H+ (molar HCl) = 0.00138 mol H+
a ) How many moles of N a O H was required to titrate KHP and any excess H C l in Trial 1?Show all work with the proper number of significant figures and units. Show numbers with short descriptors.
b) What is the molarity of the N a O H solution in Trial 1? Show all work with the proper number of significant figures and units. Show numbers with short descriptors.
a ) Moles of NaOH required to titrate KHP and any excess HCl in Trial 1= 0.001234 mol H+ (molar KHP) + 0.00015 mol H+ (molar HCl) = 0.00138 mol NaOH
b) Molarity of the NaOH solution in Trial 1= (0.00138 mol (molar NaOH) / 4.60 mL (volume NaOH)) x (1000 mL / 1 L) = 0.301 M NaOH
What is the average molarity of you sodium hydroxide solution? Report the average based on the number of trials you instructor asked you to perform.
Type the molarities of each trial and show how the average was calculated.
Average molarity of sodium hydroxide solution= 0.301 M (molarity NaOH for Trial 1) + 0.318 M NaOH (molarity NaOH for Trial 2) + 0.303 M (molarity NaOH for Trial 3) / 3 (number of trials) = 0.307 M NaOH solution
Your instructor will give you instructions as what to do with this box. It may be used for bonus points, or for taking away points (for example late submissions) or for error analysis if something went wrong during the lab. If no instructions have been provided to you, please leave this box blank.
5% = 5
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