Prelab Questions help needed!! This an advance titration lab I have to do next w
ID: 475291 • Letter: P
Question
Prelab Questions help needed!! This an advance titration lab I have to do next week but I have to answer the pre-lab questions first. I am super confused on how to do the ones down there. Please help, and please show work on you obtained the answer. Thank you very much in advance!!
1) A beaker contains 19.6812 g of water at 20 degrees celscius. Using the table below, calculate the "actual volume" of the water.
Volume occupied by 1.000 g of water weighed in air at various temperatures
Volume(mL)
2) At the end point, all of the KHP from your sample will have reacted with NaOH. How will you determine the number of moles of NaOH that reacted with the KHP?
3) Once you've determined the molarity of your NaOH soluiton, you will use it as a standard to determine the concentration of your HCl solution. How will you determine the number of moles of NaOH used in the titration with HCl?
4) Show how you will determine the number of moles of HCl that reacted with the NaOH.
Temperature ('C) volume (mL) Temperature ('C) Volume (mL) Temperature ('C)Volume(mL)
Temperature('C) Volume(mL) 11 1.0014 16 1.0021 21 1.0030 26 1.0043 12 1.0015 17 1.0022 22 1.0033 27 1.0045 13 1.0016 18 1.0024 23 1.0035 28 1.0048 14 1.0018 19 1.0026 24 1.0037 29 1.0051 15 1.0019 20 1.0028 25 1.0040 30 1.0054Explanation / Answer
1) You are given a table which shows the “actual volume” of 1.000 g of water at temperatures between 11-30C. Use the table to find out the “actual volume” of 1.000 g of water at 20C – the answer is 1.0028 mL.
Therefore, “actual volume” of 19.6812 g of water at 20C = (19.6812 g)*(1.0028 mL/1.000 g) = 19.7363 mL (ans).
2) Write down the balanced chemical reaction between KHP and NaOH:
KHP (aq) + NaOH (aq) -----> NaKP (aq) + H2O (l)
Molar mass of KHP = 204.22 g/mol.
Let us say you weighed out w1 g of KHP; therefore, moles of KHP weighed out = (w1 g)/(204.22 g/mol) = 0.004897w1 mole.
As per the balanced stoichiometric equation above,
1 mole of KHP = 1 mole of NaOH …….(1)
Therefore, 0.004897w1 mole KHP = 0.004897w1 mole NaOH.
We determine the moles of NaOH used in the standardization reaction by employing equation (1) above (ans).
3) In the titration experiment, we first standardize the NaOH solution and then use the NaOH solution to standardize the given HCl solution. We have already determined the moles of NaOH used in the reaction as 0.004897w1 (from part 2 above).
Suppose we make a 250 mL solution of the standard NaOH solution. We can calculate the molarity of the prepared NaOH solution by using the relation
Molarity = moles of NaOH used/volume of solution in L = (0.004897w1)/[(250 mL)*(1 L/1000 mL)] = 0.004897w1/0.250 mole/L.
Next, let us say we pipette out 25 mL of standard NaOH solution for titration with HCl. We can simply find out the moles of NaOH pipetted out by employing the relation
Number of moles = Molarity of solution*volume of solution pipetted out in L = (0.004897w1/0.250 mol/L)*(25 mL)*(1 L/1000 mL) = 0.0004897w1 (ans).
4) Write down the balanced reaction between NaOH and HCl:
NaOH (aq) + HCl (aq) -------> NaCl (aq) + H2O (l)
As per the balanced stoichiometric reaction,
Moles of NaOH added = moles of HCl titrated.
Therefore, moles of HCl titrated = 0.0004897w1 (ans).