Question

Đã trả lời bởi chuyên gia

Answer 1

a,

Ta có:

$n_{H^{+}} = n_{HCl} + 2 . n_{H_{2} {SO}_{4}} = 0, 5.0, 1 + 2.0, 5.0, 2 = 0, 25 (mol) n_{{OH}^{-}} = n_{KOH} + 2 . n_{Ba {(OH)}_{2}} = 0, 5.0, 1 + 2.0, 5.0, 1 = 0, 15 (mol) PTHH : H^{+} + {OH}^{-} \to H_{2} O Ta có : \frac{n_{H^{+}}}{n_{{OH}^{-}}} = \frac{0, 25}{0, 15} > \frac{1}{1} \Rightarrow H^{+} dư, {OH}^{-} hết \Rightarrow n_{H^{+} dư} = 0, 25 - 0, 15 = 0, 1 (mol) V_{dd sau pư} = 0, 5 + 0, 5 = 1 (l) \Rightarrow [H^{+} dư] = \frac{0, 1}{1} = 0, 1 (M) \Rightarrow pH sau pư = - \log 0, 1 = 1 (M) b, n_{Ba {(OH)}_{2}} = 0, 5.0, 1 = 0, 05 (mol) n_{H_{2} {SO}_{4}} = 0, 5.0, 2 = 0, 1 (mol) PTHH : {Ba}^{2 +} + {SO}_{4}^{2 -} \to {BaSO}_{4} ↓ Ta có : \frac{n_{{Ba}^{2 +}}}{n_{{SO}_{4} 2^{-}}} = \frac{0, 05}{0, 1} < \frac{1}{1} \Rightarrow n_{{BaSO}_{4}} = n_{{Ba}^{2 +}} = n_{Ba {(OH)}_{2}} = 0, 05 (mol) \Rightarrow m_{{BaSO}_{4}} = 0, 05.233 = 11, 65 (g)$

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Answer 2

a,

Ta có:

$n_{H^{+}} = n_{HCl} + 2 . n_{H_{2} {SO}_{4}} = 0, 5.0, 1 + 2.0, 5.0, 2 = 0, 25 (mol) n_{{OH}^{-}} = n_{KOH} + 2 . n_{Ba {(OH)}_{2}} = 0, 5.0, 1 + 2.0, 5.0, 1 = 0, 15 (mol) PTHH : H^{+} + {OH}^{-} \to H_{2} O Ta có : \frac{n_{H^{+}}}{n_{{OH}^{-}}} = \frac{0, 25}{0, 15} > \frac{1}{1} \Rightarrow H^{+} dư, {OH}^{-} hết \Rightarrow n_{H^{+} dư} = 0, 25 - 0, 15 = 0, 1 (mol) V_{dd sau pư} = 0, 5 + 0, 5 = 1 (l) \Rightarrow [H^{+} dư] = \frac{0, 1}{1} = 0, 1 (M) \Rightarrow pH sau pư = - \log 0, 1 = 1 (M) b, n_{Ba {(OH)}_{2}} = 0, 5.0, 1 = 0, 05 (mol) n_{H_{2} {SO}_{4}} = 0, 5.0, 2 = 0, 1 (mol) PTHH : {Ba}^{2 +} + {SO}_{4}^{2 -} \to {BaSO}_{4} ↓ Ta có : \frac{n_{{Ba}^{2 +}}}{n_{{SO}_{4} 2^{-}}} = \frac{0, 05}{0, 1} < \frac{1}{1} \Rightarrow n_{{BaSO}_{4}} = n_{{Ba}^{2 +}} = n_{Ba {(OH)}_{2}} = 0, 05 (mol) \Rightarrow m_{{BaSO}_{4}} = 0, 05.233 = 11, 65 (g)$

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