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Working with weak bases and their dissociation - finding pH

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Methylamine, CH3NH2, is a weak base that reacts according to the equation CH2NH2 + H2O <--> CH3NH3^1+ + OH^1-. The value of the ionization constant, Kb, is 5,25 x 10^-4. Methylamine forms salts such as methylammonium nitrate, (CH3NH3^1+)(NO3^1-). (b) Calculate the pH of a solution maded by adding 0.0100 mole of a solid methylammonium nitrate to 120.0 mL of a 0.225-molar solution of methylamine. Assume that no volume change occurs. (The answer i got was pH=11.15) (c) How many moles of EITHER NaOH OR HCl (state clearly which you choose) should be added to the solution in (b) to produce a solution that has a pH of 11.00? Assume that no volume change occurs.
(d) A volume of 100 mL of distilled water is added to the solution in (c). How is the pH of the solution affected? Explain.
This is from AP Chem Exam 1993 ques #1

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<br>We have the reaction, CH2NH2 + H2O &lt;--&gt; CH3NH3^1+ + OH^1-,
<br>
<br>for this, Kb = [CH3NH3+]*[OH-]/[CH3NH2]
<br>
<br>here [] represents the concentration of the respective, ions.
<br>[CH3NH2]= 0.225, BUT WE DON'T KNOW THE OTHER TWO.
<br>
<br>Now we assume that x moles of ch3nh3 undergoes dissociation. Thus equal amounts of CH3NH3 + and OH - will be formed
<br>
<br>thus, [CH3NH3+] = [OH-] = X
<br>
<br>NOW, SUB BACK. Kb = [CH3NH3+]*[OH-]/[CH3NH2] = x*x/0.225
<br>(actually we must take the denomenator as 0.225-x but the dissociation is very ...

Solution Summary

Question from AP Chem Exam 1993, explained with all mathematical steps. A discussion is also given.

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Plotting a titration curve for Ala and His

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1. Assume you have a solution containing 0.1 mol of Ala adjusted to pH = 0.5 with HCl. You begin adding 1.0 M NaOH. Plot (do not sketch) the resulting titration curve (pH vs. moles NaOH added) showing all inflection points. Show your calculations and be sure to state your assumptions.

2. Plot a similar titration curve for a solution containing 0.1 mol of His at pH = 0.5. (pKa values: Ala: alpha-COOH = 2.3, alpha-NH2 = 9.7; His: alpha-COOH = 1.8, alpha-NH2 = 9.2, Imidazole = 6.0).

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