Test Bank, Question 4.235 [ Your answer is incorrect. Try again pter 4 CHM 301 A

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Test Bank, Question 4.235 [ Your answer is incorrect. Try again pter 4 CHM 301 A student in a laboratory performed a titration to find out the exact molarity of her unknown KOH(aq) solution. She weighed out 2.465 g of KHC8H404, dissolved it in 100 mL of water and titrated it with the base The equation for the reaction is: est Bank, GrestsonThe titration showed that 42.5 mL of the base were required to neutralize the KHCgH404 sample. What is the molarity of the potassium hydroxide solution? mtsrs.oesten Use correct number of significant digits; the tolerance is +/.2% ractice Intellgent Question Attempts:1 of 5 used SAVE FOR LATER est Bank. Question utorina Question 4.11 ractice Intelligent ew Results by Study

Explanation / Answer

The molar mass of KHP, the analyte used is 204.22g/mol. So, the molarity of 100mL of KHP solution containing 2.465g of the salt is calculated as Molarity = (Weight/Molar mass)x(1000/Volume) = (2.465/204.22)x(1000/100) = 0.1207M.

Now, to find the molarity of the unknown KOH solution, we use the law of mass conservation and law of equivalence to relate to the acidic salt and base's concentration and volume as VKHPMKHP = VKOHMKOH which gives MKOH = (VKHPMKHP)/VKOH = (100x0.1207)/42.5 = 0.2840M.

Thus, it can be determined that the concentration of the unknown KOH is 0.2840M.

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