A volume of 10.50 mL of 0.1090 M NaOH solution was used to titrate a 0.699 g sam

Question

A volume of 10.50 mL of 0.1090 M NaOH solution was used to titrate a 0.699 g sample of unknown containing KHSO_4. What is the molecular mass of KHSO_4?(report answers to 4 or 5 significant figures_ __________ Tries 0/99 What is the percent by mass of KHSO_4 in the unknown? ______ Tries 0/99 In this problem what mass of sample in grams would be needed to deliver about 22.00 mL in the next trial? _____ Tries 0/99 In the second trial above, exactly 1.462 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. ______ Tries 0/99

Explanation / Answer

1) Volume of NaOH required for the titration = 10.50 mL; concentration of NaOH solution = 0.1090 M.

Therefore, moles of NaOH required for the titration = (10.50 mL)*(0.1090 mol/L) = 1.1445 mmole.

Write down the neutralization reaction as below:

NaOH + KHSO4 ---------> NaKSO4 + H2O

As per the balanced stoichiometric reaction,

1 mole NaOH = 1mole KHSO4.

Therefore, 1.1445 mmole NaOH = 1.1445 mmole KHSO4

Let M be the molar mass of KHSO4; therefore,

0.699 g/M = 1.1445 mmole = 1.1445*10-3 mole.

===> M = (0.699 g)/(1.1445*10-3 mole) = 610.747 g mol-1 610.75 g mol-1 (ans).

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