what volume of 0.812 M HCl, in milliliters, is required to titrate 1.45 g of NaOH to the equivalence point? ?
I have a good understanding of titration problems, but I'm unsure how to approach the calculation involving 1.45 grams of NaOH. Specifically, what volume of 0.812 M HCl, in milliliters, is needed to reach the equivalence point in this titration?
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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