What is the [OH^{-}] in the final solution pr | vedclass

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(D) Number of millimoles of $HCl = 20.0 	imes 0.050 = 1.0 \ mmol$. Number of millimoles of $Ba(OH)_2 = 30.0 	imes 0.10 = 3.0 \ mmol$. Since $Ba(OH)_

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$0.10$

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(D) Number of millimoles of $HCl = 20.0 	imes 0.050 = 1.0 \ mmol$. Number of millimoles of $Ba(OH)_2 = 30.0 	imes 0.10 = 3.0 \ mmol$. Since $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^{-}$,$3.0 \ mmol$ of $Ba(OH)_2$ provides $6.0 \ mmol$ of $OH^{-}$ ions. Number of $H^{+}$ ions from $HCl = 1.0 \ mmol$. Remaining $OH^{-}$ moles after neutralization $= 6.0 - 1.0 = 5.0 \ mmol$. Total volume $= 20.0 + 30.0 = 50.0 \ mL$. $[OH^{-}] = \frac{5.0 \ mmol}{50.0 \ mL} = 0.10 \ M$.

What is the $[OH^{-}]$ in the final solution prepared by mixing $20.0 \ mL$ of $0.050 \ M \ HCl$ with $30.0 \ mL$ of $0.10 \ M \ Ba(OH)_2$? $...... \ M$

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