6. What weight of sample should be taken so that the milliliters of 0.1026 M NaOH titrant will equal the % KHP in the sample?

7. Calculate the region of 0.1 % accuracy (i.e. 99.9% titration to 100.1% titration) for a solution of H+NO3 which is to be titrated with 0.1000 M Na+OH . Assume you have 10.00 m moles of HNO3 initially in 100.0 mL of solution. ANS: pH of 4.30 to pH of 9.70

8. Calculate the region of 0.1 % accuracy (i.e. 99.9% titration to 100.1% titration) for a solution of HOAc,
( MW = 60.00, pKa = 4.75) which is to be titrated with 0.1000 M Na+OH . Assume you have 10.00 m moles of HOAc initially in 100.0 mL of solution.

9. You wish to prepare 1000 mL of a buffer solution whose pH value is 10.0 by mixing together ammonia (NH3, MW = 17.00, Kb = 1.8 x 10 5) and ammonium chloride (NH4+Cl , MW = 53.49). If the "total nitrogen content" is to be no more than 0.15 M,
a) how many grams of NH4Cl are necessary
b) how many mL of conc NH3 (15.0 M) are necessary

10. Measurement of the diameter of a piece of glass tubing yields the following values (in mm.): 5.0, 5.5, 5.0, 5.5, 6.0.
a) What is your best estimate of the diameter of the tubing
b) What is the precision of this set of measurements, expressed as % RSD? (N.B. if you choose to use the range method, the appropriate k is 0.43)

Clueless:

The best way to proceed here is to set up a template, the skeleton set-up by which you determine the % analyte for any titration set of data. In this case the analyte is %KHP, so you want to set up a calculation by which you'll calculate the %KHP.

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Hints:

When you set up the template, you'll be able to put in values for everything but
I -- the sample weight (this is the unknown you seek)
II -- the ml NaOH used in the titration calculation
III -- the actual value for %KHP for the titration calculation.

It would seem there are 3 unknowns, but you are told that II and III are numerically the same. Thus they will cancel, leaving you with only the desired calculation for sample wt.

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Clueless:

This is simply the establishment of a titration curve for a strong acid being titrated by a strong base. The [H3O+] is determined "directly" up to the equivalence point.

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Hints:

If you start with 10.00 mmol of HNO3 then to reach 100.0% titration you'll need 10.00 mmol of NaOH (corresponding to the addition of 100.0 mL of NaOH titrant, as each mL brings 0.1000 mmol of NaOH). When you're 99.9% titrated you've added 9.99 mmol NaOH (99.90 mL of 0.1000 M NaOH) leaving 0.01 mmol of H3O+ in a total of 100.0 + 99.90 mL solution. When you're 100.1 % titrated you've added 10.01 mmol NaOH (100.1 mL of 0.1000 M NaOH) leaving you with 0.01 mmol OH- in a total of 100.0 + 100.0 mL solution.

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Clueless:

This is simply the establishment of a titration curve for a weak acid being titrated by a strong base. The [H3O+] is determined cannot be determined "directly" up to the equivalence point, as in a strong acid case, but must be determined, most easily, through the use of the logarithmic form of the Ka expression, the H-H Equation (except for the initial calculation which cannot use the H-H Equation).

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Hints:

Recall that at the equivalence point you'll have one species with acid/base properties in the solution (not counting water), the conjugate of HOAc. It is this conjugate, OAc- whose Kb will set the pH of the solution. This contrasts with the Strong Acid case where at the equivalence point you have NO species (other than water) present which has acid/base properties, leaving it to water to set the pH for this case through a Kw calcn.

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Clueless:

This is a buffer determination problem. You are told that it is a 2-component buffer, NH3 and NH4Cl. You need to decide which is the acid and which is the conjugate so that you can use the Henderson-Hasselbalch equation to calculate the ratio of these two components which will fix the pH of the solution to the desired value.

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Hints:

Here even though you'll probably be given the Kb for NH3, it's probably better to view this in terms of NH4+ being the acid and using its pKa (which is easily obtained from the pKb for NH3). The rest of the procedure is straightforward:
A. Write the H-H eqn, specifying the formulas of the acid and conjugate.
B. Solve the H-H eqn to determine the ratios of the acid and conjugate. At the moment you have 2 "unknowns", the conc of acid, and the conc of conjugate.
C. From the information that the "total carbonate concentration" is to be no more than 0.1500 M, you can set one of your unknowns to the classic value X. The other unknown will then be (0.1500 - X). Substituting these into the value for the ratio of these two will permit you to calc the concentration of the acid and conjugate.
D. You now are faced with preparing a solution of, in this case, 750 mL for a formula whose concentration you've just determined in step 5 above. This is a straightforward chem 111 problem set calculation (P.S. 130) with the added bonus that one of calculations will be for a solid (NH4Cl) while the other will involve the dilution of a concentrated liquid (NH3).

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Clueless:

This is a straightforward statistics calculation whereby you determine the central tendency of a set of measurements and the quality associated with these measurements.

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Hints:

Part a) While a case can be made for using the MEDIAN in this kind of work, most people would automatically calculate the MEAN.

Part b) Recall the definition %RSD , a unitless measure of the quality of your measurements. Also recall, that it is simpler to use the Range Method to determine sy, the sample standard deviation, than the Sum of Squares Method (the price that is paid is that sy is generally a numerically larger value when determined using the Range Method). This leads to a larger "Confidence Limits Box" for the same level of assurance (for us in this course 95%).

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