Help Please :) NaOH

gh1991

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13 years ago

For each of the following circumstances indicate where the calulated molarity of NaOH would be lower, higher, or unaffected. Explain your answer:
a) the inside of the pipet used to transfer the standard HCl solution was wet with water.
b) You added 40 mL of water to the titration flask rather than 25 mL
c) the buret, wet with water was not rinsed with NaOH solution before filling the buret with NaOH solution,
d) 5 dropes of phenolphthalein were added to the solution to be titrated rather than 3 drops.

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padre

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13 years ago

mols HCl = L x M
If pipet had water, then we transferred too little HCl; i.e., L is too small which will make mols HCl smaller than we think. When we titrate with NaOH,
M NaOH = mols/L. We have fewer mol than we think we have, that means L NaOH will be smaller, and mols/smaller number means a higher M that we calculate for NaOH.
The others are done the same way.

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duddy

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Staff Member

13 years ago

For each of the following circumstances indicate where the calulated molarity of NaOH would be lower, higher, or unaffected. Explain your answer:

a) the inside of the pipet used to transfer the standard HCl solution was wet with water.

If there was water in the pipette, then less HCl would be transferred. However, NaOH reacts with water to form hydroxide ions. If there is HCl in the pipette as well, it will react with NaOH, thereby causing its molarity to lower.

b) You added 40 mL of water to the titration flask rather than 25 mL

Same scenario as before, the more water, the more NaOH will dissociate, decreasing the molarity.

c) the buret, wet with water was not rinsed with NaOH solution before filling the buret with NaOH solution,

Decrease.

d) 5 dropes of phenolphthalein were added to the solution to be titrated rather than 3 drops.

Unaffected.

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