Calculate the [H + ] of a 0.50 M solution of methylammonium bromide, CH 3 NH 3 Br. The K b of methylamine, CH 3 NH 2 , is given as 4.4 x 10 -4 .

Calculate the H+ ion concentration of a 0.50 molar Methylammonium Bromide solution. Alright. So here, methyl ammonium bromide, it may not seem obvious, but it is an ionic compound. It's composed of a positive ion in the form of Methylammonium ion, plus a negative ion, which is the bromide ion. Remember, positive amines like this are weak acids. We can also say that they are acidic ions. For negative ions, you add an H plus to them to create an acid. The acid that we create is hydrobromic acid, which is a strong acid. Remember, if you create a strong acid that means the negative ion is neutral, and since it's neutral we can ignore it. So we're gonna take the methyl ammonium ion, which is our weak acid, and have it react with water. So it is aqueous since it's an acid, it reacts with water which will be a liquid. Since it is an acid it's going to donate an H+ ion to the water which acts as the base. The H plus comes from the nitrogen atom itself, so we'd have CH3NH2 aqueous, so that's the conjugate base, plus H3O+ ion. Since it's a weak acid, we're gonna set up an ICE chart. So we have initial change equilibrium, we're gonna say here in an ICE chart, we're gonna use the initial concentration of our weak acid. In an ICE chart we ignore solids and liquids. The products initially are 0. Remember we lose reactants in order to make products. So bring down everything, s x, plus x. Now, we need to set up our equilibrium expression. And since this is a weak acid it's going to utilize Ka. So we need to find the Ka of Methylammonium ion. Now, if you look in your textbook, you won't find the Ka for this structure. And in fact, if you're looking at a typical Ka or Kb chart, you'll see a vast majority of your examples are neutral species. That's because you won't find the K of this, but you will find the KB of its conjugate base. Its conjugate base is CH3 NH2. So here look up Methylamine. If you do that you'll see that the kb is 4.4 times 10 to the negative 4. Now that we have the kb, we can find Ka, because remember k a times k b equals k w. K a equals k w divided by k b, so that would be 1.0 times 10 to the negative 14, divided by 4.4 times 10 to the negative 4. This comes out to give me 2.27 times 10 to the negative 11 for my ka value. Now, here we're going to say Ka equals products over reactants, so we're gonna say k equals my 2 products multiplied times each other, divided by my weak acid form. Water is a liquid so water will be ignored. Plug in the numbers that we have. We have 2.27 times 10 to the negative 11 equals, at equilibrium both of my products are x, multiplied together that would be x squared, divided by 0.50 x. Now this -x portion here we can determine if it can be ignored or not. We do that by utilizing our 500 approximation method. Now remember that your 500 approximation method we take the initial concentration, and we divide it by the k a of our weak acid. If we get a number greater than 500, then we can ignore the minus x. So when I do initial concentration divided by my k value I get 2.20 times 10 to the 10, a number much greater than 500 which means I can ignore this minus x here. By ignoring it I can avoid the quadratic formula. So we're gonna cross multiply these 2, so x squared equals 1.135 times 10 to the negative 11. Take the square root of both sides here, when we do that x equals 3.37 times 10 to the negative 6 molar. We're looking for the H+ ion concentration. Remember, H+ ion concentration is the same thing as H three O+ ion concentration. And at equilibrium, H3O plus equals x. So we just figured out what the H plus concentration is. It's this 3.37 times ten to the negative six molar.

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17. Acid and Base Equilibrium

pH of Weak Acids

General Chemistry

17. Acid and Base Equilibrium

pH of Weak Acids

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Multiple Choice

Calculate the [H⁺] of a 0.50 M solution of methylammonium bromide, CH₃NH₃Br. The K_b of methylamine, CH₃NH₂, is given as 4.4 x 10^-4.

0.50 M

1.14 × 10⁻¹¹ M

3.4 × 10⁻⁶ M

2.97 × 10⁻⁹ M

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Verified step by step guidance

Identify the chemical species involved: Methylammonium bromide (CH3NH3Br) dissociates in water to form CH3NH3+ and Br-. The CH3NH3+ ion is the conjugate acid of methylamine (CH3NH2).

Understand the relationship between Kb and Ka: The Kb value given is for methylamine (CH3NH2). To find the Ka for its conjugate acid (CH3NH3+), use the relation Ka * Kb = Kw, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C).

Calculate Ka: Rearrange the equation to find Ka = Kw / Kb. Substitute the values: Ka = (1.0 x 10^-14) / (4.4 x 10^-4).

Set up the equilibrium expression for CH3NH3+ dissociation: CH3NH3+ ⇌ CH3NH2 + H+. The initial concentration of CH3NH3+ is 0.50 M. Let x be the change in concentration at equilibrium, where [H+] = x.

Use the expression for Ka: Ka = [CH3NH2][H+]/[CH3NH3+]. Substitute the equilibrium concentrations: Ka = (x)(x)/(0.50 - x). Assuming x is small compared to 0.50, simplify to Ka ≈ x^2/0.50. Solve for x to find [H+].