III. Questions . 1.During the titration several milliliters of deionized water a
ID: 994207 • Letter: I
Question
III. Questions.
1.During the titration several milliliters of deionized water are added to the Erlenmeyer flask containing the KHP solution. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
2.A 25.00 mL volumetric pipet is rinsed with deionized water. Several drops of water remain in the pipet. It is then used to deliver a 25.00 mL aliquot of the KHP solution into the Erlenmeyer flask in which the sample will be titrated. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
3.A buret is cleaned and rinsed with deionized water. Some water remains in the buret. It is then filled with the NaOH solution which is being standardized. How will this affect the calculated concentration of NaOH (too high, too low or unchanged?) Explain clearly.
4.During the preparation of the KHP solution, some KHP is spilled on the counter. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
Explanation / Answer
1.During the titration several milliliters of deionized water are added to the Erlenmeyer flask containing the KHP solution. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
Ans: No effect on the calculated NaOH concentration
You pipette out a known volume of the KHP solution (say 25 mL) in the flask and this volume contains certain moles of KHP (depending on the strength of the solution). Addition of any deionized water will not change the number of moles of KHP in the flask which are going to react with NaOH that you add from the burette. Thus the volume of NaOH required for neutralization remains the same irrespective of the volume of deionized water added. While calculating the normality of NaOH using the normality equation, you will use the volume of KHP which you pipetted out initially (i.e 25 mL). Thus the result will remain unchanged.
NKHPVKHP = NNaOH VNaOH
Or, NNaOH = (NKHPVKHP)/ VNaOH
2.A 25.00 mL volumetric pipet is rinsed with deionized water. Several drops of water remain in the pipet. It is then used to deliver a 25.00 mL aliquot of the KHP solution into the Erlenmeyer flask in which the sample will be titrated. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
Ans: Calculated concentration of NaOH will be higher than true value
In this case the expelled volume of KHP will be less than the intended volume (say 25 mL, this intended volume is the one that you use in the normality equation during calculations). Thus the volume of NaOH required for neutralization would be less than the actual value. Hence the calculated strength would be higher than the true strength of NaOH (see equation above, if VNaOH decreases, NNaOH increases)
3.A buret is cleaned and rinsed with deionized water. Some water remains in the buret. It is then filled with the NaOH solution which is being standardized. How will this affect the calculated concentration of NaOH (too high, too low or unchanged?) Explain clearly.
Ans : Calculated concentration of NaOH will be lower than true value
In this case the NaOH solution will get diluted with the water already present in the burette. Thus when you do the titration, you will require a larger volume of NaOH to neutralize a given volume of KHP. This will give you a lower value for NNaOH upon calculation.
4.During the preparation of the KHP solution, some KHP is spilled on the counter. How will this affect the calculated concentration of NaOH (too high, too low, or unchanged?) Explain clearly.
Ans : Calculated concentration of NaOH will be higher than true value
If some KHP is spilled on the counter during solution preparation, that means fewer moles of KHP (than intended/calculated) are present in the solution per unit volume. In other words, the actual normality of the KHP solution is less than the calculated normality. Thus the same volume of KHP solution will require less volume of NaOH for neutralization. This will give you a higher normality of NaOH.