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so in our lab, we measured KHP and titrated it under sodium hydroxide. our goal was to make the solution tainted pink. 
for each of the following scenarios, state how this error would effect your calculated concentration of sodium hydroxide. would the concentration be artifically7 high or artificially low if you had actually made this experimental mistake? explain.
1. while you were measuring out your KHP, some of the powder missed the flask and fell onto the scale. 
2. you were looking up at teh burette. you thought it read 0.00mL as your initial colume, but it actually read 0.12mL (assume your final volume reading was accurate)
3. some of the sodium hydroxide you released from teh burette hit the side of the burette instead of dropping down and reacting with rthe KHP.
4. your titrated solution was bright pink.

Chem Man · Accepted Answer

KHP + NaOH ----->  NaKP + H2O
1 mol KHP reacts with 1 mol NaOH

1.
The recorded mass of KHP would be higher than the mass in the flask. With less KHP, less volume of NaOH would be required to react. The effect is to make the calculated molarity of NaOH too high. The calculated mol of NaOH would be too high.
M = mol / L  If the mol is too high, the M is too high.

2.
The recorded volume of NaOH is too high. The volume of NaOH used is the final burette reading minus the initial reading. If the initial reading is too low (0.00), the calculated volulme used is to high. This causes the calculated M of NaOH to be too low.  M = mol / L, if L is too large, M is too low.

3.
The recorded volume of NaOH added would be too great because you had to add extra NaOH to make up for the base that didn't make it into the reaction. M = mol / L. When the recorded V is too high, the M will be too low.

4.
If the solution is too pink, it means that an extra drop (or more) of NaOH was added beyond the end point. This causes the final burette reading to be too large and the overall volulme of NaOH used to be too large. M = mol / L. If the volume is too large, the calculated M will be too low.

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help with chemistry?

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