In: Chemistry

Calculate the concentration of all species present (H2SO4, HSO4^-, H^+, SO4^-2) in a 0.285 M H2SO4 solution. The diprotic acid equilibrium is :

H2SO4 <--> H^+ + HSO4^- Ka = very large

HSO4^- <---> H^+ + SO4^-2 Ka = 1.2 x 10^-2

Please explain why and when we use +x and -x. I am very confused on this.

H_{2}SO_{4}
H^{+} + HSO_{4}^{-}
K_{a1} = very large (about 1 x 10^{3})

HSO_{4}^{-}
H^{+} + SO4^{-2} K_{a2} = 1.2
x 10^{-2}

Given, concentration of H_{2}SO_{4} = 0.285
M

For the first equibrium,

H_{2}SO_{4}
H^{+} + HSO_{4}^{-}

let x parts of H_{2}SO_{4} get dissociated

So, from Initial change and equilibrium table or ICE Table

Particulars |
H_{2}SO_{4} |
H^{+} |
HSO_{4}^{-} |

Initial | 0.285 | 0 | 0 |

change | -x | +x | +x |

equilibrium | 0.285-x | +x | +x |

-x implies that x part of the reactant gets dissociated or lost
from it. Here H_{2}SO_{4} loses x part to give x
parts of H^{+} and HSO_{4}^{-}
each

+x implies that x part of the products formed . Here x parts of
both H^{+} and HSO_{4}^{-} are
formed from x part of H_{2}SO4. let's say 1 unit gets
dissociated from H_{2}SO4. So, 1 unit each of both
H^{+} and HSO_{4}^{-} are
formed.

+x is used when reactant gets dissociated and -x is used products are formed

now, since H_{2}SO_{4} is a strong acid, hence
its dissociation is complete,

i.e. 0.285-x = 0 or x = 0.285 M

For second reaction,

HSO_{4}^{-}
H^{+} + SO4^{-2}

let y parts of HSO_{4}^{-} gets dissociated
now

From ICE table,

Particulars |
HSO_{4}^{-} |
H^{+} |
SO4^{-2} |

Initial | x | x | 0 |

Change | -y | +y | +y |

Equilibrium | x-y | x+y | +y |

K_{a2} = [H^{+}][SO4^{-2}
]/[HSO_{4}^{-}]

1.2 x 10^{-2} = (x+y)(y)/(x-y)

= (0.285+y) y / (0.285-y)

or, y^{2} + 0.297
y-0.00342 = 0

y= [-0.297 {0.297^{2}
- (4*1*(-0.00342))}^{1/2}]/2

= (-0.297 0.319) / 2

or, y = -0.616 or 0.022

since, concentration cannot be negative, y= 0.022 M

Hence, at equilibrium,

[H_{2}SO_{4}] = 0 M,

[H^{+}] = x+y = 0.285 + 0.022 = 0.307 M

[HSO_{4}^{-} ] = x-y = 0.285 -0.022 = 0.263
M

[SO4^{-2}] = y = 0.022 M

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[H2GeO3]______ M
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Calculate the pH of a 9.06×10-3 M solution of
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Part A
Calculate the pH and the concentrations of all species present
in 0.16 M H2SO3. (Ka1 = 1.5×10−2, Ka2 =
6.3×10−8)
Express your answer to three significant figures and include the
appropriate units.
pH =
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Up
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Part B
Calculate the concentration of H2SO3 in solution.
Express your answer to two significant figures and include the
appropriate units.
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Incorrect; Try Again; 5 attempts remaining
Part C
Calculate the concentration...

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