When you post a problem like this you should tell us what you are using for pKa or pKb; otherwise we can come up with the same answers you may have since tables containing those pK values are not consistent from text to text.
You have 0.445 x 150 mL = approx 67 millimols of the base that that's an estimate and you should do it more carefully than that.
.......CH3NH2 + HCl ==> CH3NH3^+ + Cl^-
I.......67...............0
add..............x..............
C........-x.....-x.......x
E......67-x......0........x
Now substitute the E line into the Henderson-Hasselbalch equation and solve for x = millimols HCl to be added.
Then mmols HCl = M x mL. You know mmols and M, solve for mL.
An aqueous solution contains 0.445 M ethylamine (C2H5NH2). How many mL of 0.226 M hydrochloric acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 10.200
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1 answer