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MrRa [10]
3 years ago
5

Calculate the [H3O+] of 0.35 M solution of benzoic acid HC7H3O2

Chemistry
1 answer:
solmaris [256]3 years ago
5 0

Answer:

[H_3O^+]=4.7x10^{-3}M

Explanation:

Hello there!

In this case, since the ionization of benzoic acid is:

HC_7H_3O_2+H_2O\rightleftharpoons C_7H_3O_2^-+H_3O^+

Whereas the corresponding equilibrium expression is:

Ka=\frac{[C_7H_3O_2^-][H_3O^+]}{[HC_7H_3O_2]}

Now, we insert the acidic equilibrium constant and the reaction extent x, to write:

6.3x10^{-5}=\frac{x^2}{0.35M}

Thus, by solving for x, we obtain:

x=\sqrt{6.3x10^{-5}*0.35}\\\\x=4.7x10^{-3}M

Which is also:

x=4.7x10^{-3}M

Best regards!

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3 years ago
Read 2 more answers
Calculate the molar solubility of ca(io3)2 in each solution below. the ksp of calcium iodate is7.1 × 10−7.
e-lub [12.9K]
In order to find the answer, use an ICE chart:

Ca(IO3)2...Ca2+......IO3- 
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x 
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
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<span>x = molar solubility = 5.6 x 10</span>⁻³ M

The answer is 5.6 x 10 ^ 3 M. (molar solubility)
5 0
3 years ago
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