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SVETLANKA909090 [29]
3 years ago
15

If aqueous solution of lead (ll) nitrate and sodium sulfate, which insoluble precipitate is formed

Chemistry
1 answer:
forsale [732]3 years ago
7 0

Answer:

Lead(II) sulfate

Explanation:

This looks like a double displacement reaction, in which the cations change partners with the anions.

The possible products are

Pb(NO₃)₂ (aq)+ Na₂SO₄(aq) ⟶PbSO₄(?) + 2NaNO₃(?)

To predict the product, we must use the solubility rules. Two important ones for this question are:

  1. Salts containing Group 1 elements are soluble.
  2. Most sulfates are soluble, but PbSO₄ is an important exception.

Thus, NaNO₃ is soluble and PbSO₄ is the precipitate.

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Determine the oh of a solution that is 0.220 m in hco3
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Answer : The hydroxide ion concentration of a solution is, 5\times 10^{-14}

Explanation :

As we know that HCO_3^- dissociates in water to give hydrogen ion (H^+) and carbonate ion (CO_3^{2-}).

As, 1 mole of HCO_3^- dissociates to give 1 mole of hydrogen ion (H^+)

Or, 1 M of HCO_3^- dissociates to give 1 M of hydrogen ion (H^+)

So, 0.200 M of HCO_3^- dissociates to give 0.200 M of hydrogen ion (H^+)

Now we have to calculate the hydroxide ion concentration.

As we know that:

[H^+][OH^-]=1\times 10^{-14}

0.200\times [OH^-]=1\times 10^{-14}

[OH^-]=5\times 10^{-14}

Therefore, the hydroxide ion concentration of a solution is, 5\times 10^{-14}

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