pH (1 Viewer)

tennille

...
Joined
Nov 2, 2003
Messages
3,539
Location
Sydney
Gender
Female
HSC
2004
Hey, I'm really having trouble with a question I received in a practice paper. Could anyone help?

18g of H2SO4 is completely dissolved in 250mL of distilled water. 3g of Mg is then added to this solution. Calculate the pH of the solution after the Mg has completely reacted with the acid.

I'd really appreciate your help.

Thanks.
 

Xayma

Lacking creativity
Joined
Sep 6, 2003
Messages
5,953
Gender
Undisclosed
HSC
N/A
Mg<sub>(s)</sub>+H<sub>2</sub>SO<sub>4(aq)</sub>------>MgSO4<sub>(aq)</sub>+H<sub>2(g)</sub>

n<sub>Mg</sub>=m<sub>Mg</sub>/M<sub>Mg</sub>
=3/24.3
=0.1234567mol (yes I rounded to 1 dp above purely to get that and ignored rounding that off but it doesn't affect the final answer)

Therefore 0.1234567mol of H<sub>2</sub>SO<sub>4</sub> was removed.

n<sub>H<sub>2</sub>SO<sub>4</sub>(i)</sub>=18/98.072
=.1835
&there4;
n<sub>H<sub>2</sub>SO<sub>4</sub>(f)</sub>=.0601
n<sub>H<sup>+</sup></sub>=2n<sub>H<sub>2</sub>SO<sub>4</sub>(f)</sub>
=.1202
c=n/v
=.1202/.25
=.4808
pH=-log<sub>10</sub> .4808
=0.32
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top