Map General Chemistry 4th Edition University Science Books presented by Sapling

Question

Map General Chemistry 4th Edition University Science Books presented by Sapling Learning Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.100 M pyridine, CsHsN(aq) with 0.100 M HBr(aq) Number (a) before addition of any HBr 9.12 Number (b) after addition of 12.5 mL of HBr 5.23 With equal concentrations of monoprotic titrant and analyte, the equivalence point would occur when the added volumes are equal (25.0 mL of HBr added) (c) after addition of 21.0 mL of HBr0 So at point (c), we are 21.0/50.0-84.0% of the way to the equivalence point. If 84.0% of the base has reacted, then 16.0% remains, and 84.0% of the conjugate base has been formed so [CSH5NH yICsH5NH]84.0/16.0 Number (d) after addition of 25.0 mL of HBr 3.24 NumberFind the pOH using the Henderson-Hasselbalch equation and convert to pH for the final answer (e) after addition of 30.0 mL of HBr C H NH C,H,N Previous Give Up &View; Solution Try Again NextExit

Explanation / Answer

number of moles of Pyridine = 25 * 0.1/1000 = 2.5 *10^-3 moles

pOH = 1/2(pKb + log C)

pOH = 1/2(8.75 + log(2.5*10^-3))

pOH = 3.074

pH = 14 - 3.074 = 10.926

number of moles of HBr = 12.5 * 0.1/1000 = 1.25 * 10^-3 moles

pOH = pKb + log[salt]/[base]

pOH = 8.75 + log((1.25 * 10^-3)/((2.5*10^-3)-(1.25*10^-3)))

pOH = 8.75

pH = 14 - 8.75 = 5.25

number of moles of HBr = 21 * 0.1/1000 = 2.1 * 10^-3 moles

pOH = pKb + log[salt]/[base]

pOH = 8.75 + log((2.1 * 10^-3)/((2.5*10^-3)-(2.1*10^-3)))

pOH = 9.47

pH = 14 - 9.47 = 4.53

numberof moles of HBr = 25 * 0.1/1000 = 2.5 * 10^-3 moles

since number of moles of Pyridine = number of moles of HBr then pH of the resulting solution is equal to 7

number of moles of HBr = 30 * 0.1/1000 = 3 * 10^-3 moles

remaining moles of HBr = 3 * 10^-3 - 2.5 * 10^-3 = 5 * 10^-4 moles

pH = -log(H+)

pH = -log(5 * 10^-4)

pH = 3.301

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