Author Topic: Back titration (Read 523 times) Tweet Share

Studyinghard · « **on:** January 30, 2010, 02:15:07 pm »

A 50 ml volume of 0.10M nitric acid is mixed with 60 mL of 0.10M calcium hydroxide solution. What volume of 0.050M sulfuric acid is required to neutralise the solution ?

Thanks

fady_22 · « **Reply #1 on:** January 30, 2010, 02:32:19 pm »

EQUATIONS:
$Ca(OH)_{2}+2HNO_{3} \rightarrow 2H_{2}O+Ca(NO_{3})_{2}$
and $Ca(OH)_{2} +H_{2}SO_{4} \rightarrow CaSO_{4}+ 2H_{2}O$

Calcium hydroxide is obviously in excess in the first equation.
$n(HNO_{3})=cV=.1*0.05=0.005mol$
$n(Ca(OH)_{2} reacted)=\frac{n(HNO_{3})}{2}=0.0025 mol$
and $n(Ca(OH)_{2} added)= cV=.06*.1=0.006mol$

Therefore $n(Ca(OH)_{2})in excess=0.006-0.0025=0.0035mol$
$N(H_{2}SO_{4})=n(Ca(OH)_{2})=0.0035 mol$

Finally, $v(H_{2}SO_{4})=\frac{n}{c}=\frac{0.0035}{0.05}=0.07L=70 mL$

vexx · « **Reply #2 on:** January 30, 2010, 02:39:20 pm »

n(HNO3)= 0.05 x 0.10 = 0.005 mol
n(Ca(OH)2)= 0.06 x 0.10 = 0.006 mol

Equation:
2HNO3 + Ca(OH)2 --> Ca(NO3)2 + 2H20

Since 2 x Mol of HNO3 reacts with 1 x Mol of Ca(OH)2 we know that Ca(OH2)2 is in excess.

n(HNO3)/2= 0.0025 (amount of Ca(OH)2 reacted)

then the amount of excess is 0.006 - 0.0025 which is 0.0035 mol.

Then in the second reaction, sulfuric acid reactions on a 1:1 ratio with Ca(OH)2.

Therefore the amount of mol needed to neutralise is 0.0035mol

V(H2SO4)=0.0035mol/0.05 M
= 0.07 L required.

ahh fady beat me aha i must have begin typing this just as he was sending oh well. here's a slightly different method .__.

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Author Topic: Back titration (Read 523 times) Tweet Share

Studyinghard

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