acid-base titrations

For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case.

a.the inside of the pipet used to transfer the standard HCI solution was wet with water.
b.You added 40mL of water to the titration flask rather than 25mL
c.The buret, wet with water, was not rinsed with NaOH solution before filling the buret with NaOH solution.
d.Five drops of phenolphthalein were added to the solution to be titrated rather than 3 drops.

arent they all supposed to be affected but the last one?

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  1. b won't make a difference. One more of those listed won't make a difference. The others will. The problem asks you, not if they will be affected, but if the results will be high or low.

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  2. Ohhh, so which one would be lower, higher, or unaffected. Can you pleaseeeeeeeeeeeeee explain

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  3. For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case.

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