A volume of 12.47 mL of 0.1080 M NaOH solution was used to titrate a 0.583 g sample of unknown containing K2HPO4.

What is the percent by mass of K2HPO4 in the unknown?

In this problem what mass of sample in grams would be needed to deliver about 23.40 mL in the next trial?

In the second trial above, exactly 1.082 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 mL, predict what the final buret reading be.

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  1. K2HPO4 + NaOH ==> K2NaPO4 + H2O

    mols NaOH = M x L = ?
    Using the coefficients in the balanced equation convert mols NaOH to mols K2HPO4.
    Now grams K2HPO4 = mols x molar mass and
    % K2HPO4 = (grams K2HPO4/mass sample)*100 = ?

    For #2, plug in 0.1080 for M and 0.02340 for L and solve for mols, then use the next formula I wrote to calculate grams (g = mols x molar mass).

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