what volume of 0.812 M HCl, in milliliters, is required to titrate 1.45 g of NaOH to the equivalence point? ?

I mostly understand titration problems, but I'm not sure what to do with the 1.45 grams of NaOH

2 Answers

  • the equation is

    HCl + NaOH ------- NaCl +H2O

    HCl and NaOH are in the ratio 1:1

    mol of NaOH in 1.45g=1.45/40=0.03625mol

    therefore this is also the mol of HCl required to react with 1.45g of NaOH

    mol= concentration *volume

    volume of HCl required=0.03625/0.812=0.0446Litres to ml

    0.0446*1000=44.6millilitres

  • i think that for the time of a effective acid/base reaction the quantity of the acid is the comparable because of fact the quantity of the backside. Use M1V1=M2V2 (a million.500mol/liter)(0.03220 liter)= (a million.420 mol/L)(X). X is the unknown quantity. sparkling up for x. X= 0.0340 L=34.00 mL

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