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Question

: onv2 | Online teachin × + v nment ty.do?locatot assignment-takeatake Assignmentsessiontocatorassgrment-take M References Use the References to access important values if needed for this question. A student is asked to standardize a solution of potassium hydroxide He weighs out 0.952 g potassiun hydrogen phthalate (KHC H404, treat this as a monoprotic acid). It requires 28.0 mL of potassium hydroxide to reach the endpoint A. What is the molarity of the potassium hydroxide solution?M This potassium hydroxide solution is then used to titrate an unknown solution of hydrobromic acid B. If 11.6 mL of the potassium hydroxide solution is required to neutralize 17.0 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution? Retry Entire Group more group attempts remaining Submit Answer 0

Explanation / Answer

(A)

Amount of potassium hydrogen phthalate = 0.952 g

No. of moles of potassium hydrogen phthalate = (0.952 g) / (204.22 g/mol) = 0.004662 moles

Thus, 0.004662 moles of potassium hydrogen phthalate is neutralized with 0.004662 moles moles of potassium hydroxide

So then no of moles of potassium hydroxide = 0.004662 moles

Volume of potassium hydroxide = 28 mL = 0.028L

The molarity of potassium hydroxide = (0.004662 moles)/ (0.028L) = 0.1664 M (Ans)

(B)

Molarity of potassium hydroxide, M1 = 0.1664 M

Volume of potassium hydroxide, V1 = 11.6 mL = 0.0116 L

Molarity of hydrobromic acid, M2 = ?

Volume of hydrobromic acid, V1 = 17 mL = 0.017 L

We know that

M1V1 = M2V2

M2 = M1V1 / V2

M2 = (0.1664 x 0.0116)/ 0.017 = 0.1135 M (Ans)

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