Add 0.40 g of NaOH to 50.0 mL of 0.10 M HCOOH, and find approximate pH. No volume change, Ka HCOOH=1.8x10^-4.

I recomputed this and I get pH 5. Right?

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  1. Not even close.
    HCOOH + NaOH ==> HCOONa + H2O

    I think you don't know what you have in the solution. Answer the following. It should lead you to the correct answer.

    How many moles HCOOH initially?
    How many moles NaOH initially?
    How many moles HCOONa are produced?
    Which reagent is in excess? my how many moles?
    What is the moles/L (the concn) of the excess reagent?
    What is the pH?

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