PART B A 50.0-\ m mL volume of 0.15 \ m M \ m HBr is titrated with 0.25 \ m M \

Question

PART B A 50.0- m mL volume of 0.15 m M m HBr is titrated with 0.25 m M m KOH. Calculate the m pH after the addition of 20.0mL of m KOH. Express your answer numerically. Part C A 75.0- m mL volume of 0.200 M m NH_3 (K_{ m b}=1.8 imes 10^{-5}) is titrated with 0.500 M m HNO_3. Calculate the m pH after the addition of 15.0mL of m HNO_3. Express your answer numerically. Part D A 52.0- m mL volume of 0.35 M m CH_3COOH (K_{ m a}=1.8 imes 10^{-5}) is titrated with 0.40 M m NaOH. Calculate the m pH after the addition of 25.0mL of m NaOH. Express your answer numerically.

Explanation / Answer

we apply equality of moles for getting the remainder of acid or base so A. n1v1 = n2v2 n1v1 = 50*0.15 = 7.5 n2v2 = 0.25 * 20 = 5 so HBr is in access so access HBr = 7.5-5 = 2.5 mMoles so concentration of [HBr] = 2.5 / (50+20) = 0.0357 [H]=0.0357 pH = - log [H] = 1.447 B) same approach as above find the remainder of acid or base and apply the pH as - log[H] hope you got the concept now. thanks

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