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What is the pH of the solution formed when 0.040mol NaOH(s) is added to 2.00L of 0.020M HCl?
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What is the pH of the solution formed when 0.040mol NaOH(s) is added to 2.00L of 0.020M HCl?
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The ka expression for HTe- is?
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Answers (2)
OH! I understand now. Thanks! I didn't understand why the answer was pH 7 before.
So initial mol NaoH= 0.040mol and initial HCl= (.02M)(2.00L)= 0.04mol I don't understand what to do now.
