This is a multi-part question related to the titration of 1000 ml of 0.10 M NH4l

Question

This is a multi-part question related to the titration of 1000 ml of 0.10 M NH4l with 5.0 M KOH. How many ml of 5.0 M KOH will be required to reach the end point? In the space provided below, sketch the curve for the titration of 1000 ml of 0.10 M NH4I will 5.0 M KOH. Show the pH at the midpoint of the titration, and be sure to show the volume of KOH required to reach the endpoint of the titration. Calculate as many points as you need, including both the beginning and ending pH. Be sure to label your axes properly. Hint: For the beginning and ending pH. you do not have a buffer (need both of the acid/base conjugate pair). But for intermediate points you can use the Henderson-Hasselbalch equation: pH = pKa + log [base/acid]

Explanation / Answer

choose +x to be in the direction of motion (ie up the slope)
choose +y to be in the direction of the normal force
then

Fx = 0

P2 - P1 - Wx = 0

P2 - P1 - W sin 20 = 0

P2 - P1 = mg sin 20

Fy = 0

N - Wy = 0

N = W cos 20

N = mg cos 20

N = (16kg)(9.8m/s^2) cos 20

N = 147.3 N

and

P1 = N

P1 = (0.4)(147.3 N)

P1 = 58.94 N

so

P2 = P1 + mg sin20

P2 = 58.94 N + (16kg)(9.8m/s^2)sin 20

P2 = 112.57 N

so

P2 = 113 N

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