Calculate mass of NaN 3 that needs be added to 1.8 L of 0.35 M HN 3 in order to make a buffer with a pH of 6.5. K a of hydrazoic acid is 1.9 × 10 −5 .

Here we're told to calculate the mass of sodium azide that needs to be added to 1.8 liters of 0.35 molar of hydrazoleic acid in order to make a buffer with a pH of 6.5. Here we're told that the k of hydroxylic acid is 1.9 times 10 to the negative 5. Alright. So we're dealing with a buffer, so let's just start out with a buffer equation. And since we're dealing with Ka, we're gonna say pH equals pKa plus log of conjugate base over weak acid. In this case the conjugate base would be the sodium azide that we have here, and the weak acid will be the hydrazoic acid. Let's plug in the information given to us. We know the pH is 6.5. Now p k a is just the negative log of k a. K? So we're gonna take the negative log of the 1.9 times 10 to the negative 5. So that comes out to be 4.72. Now they're giving us the volume and molarity of our weak acid. Remember that that moles equals liters times molarity. The word of, when it's in between two numbers, means multiply. So when you multiply 1.8 times 0.35, that's going to give us the moles of hydrazoic acid. So that's going to give me 0.63 moles of hydrazolic acid. Now, we don't know the amount of our conjugate base, that's what we need to find. So let's just try to isolate that one variable. So we're gonna subtract 4.72 from both sides. When I do that, I'm gonna get 1.78 equals now this equals log of x over 0.63. Take the inverse log of both sides, so that's going to become 10 to the 1.78 equals, we take the inverse log of this side, log just drops out, so it equals x over 0.63. Cross multiply these together and that'll help you isolate x, which will equal the moles of your conjugate base. So X here would equal 37.9638 moles of sodium azide divide by 1 mole of sodium azide, And when you calculate the the molar mass of sodium azide, it comes out to be 68 oh, actually, 65 point 0 2 grams of sodium azide. Moles cancel out. We're gonna get at the end is we're gonna get 2.5 times 10 to the 3 grams of sodium azide. So this would be our final answer.

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10. Acids and Bases

Henderson-Hasselbalch Equation

GOB Chemistry

10. Acids and Bases

Henderson-Hasselbalch Equation

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Multiple Choice

Calculate mass of NaN₃ that needs be added to 1.8 L of 0.35 M HN₃ in order to make a buffer with a pH of 6.5. K_a of hydrazoic acid is 1.9 × 10⁻⁵.

1.4 g NaN₃

1.0 × 10⁻² g NaN₃

2.5 × 10³ g NaN³

3.55 g NaN³

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Verified step by step guidance

First, understand that a buffer solution consists of a weak acid and its conjugate base. In this case, the weak acid is HN3 and the conjugate base is NaN3, which dissociates to provide N3⁻ ions.

Use the Henderson-Hasselbalch equation to find the ratio of the concentrations of the conjugate base (N3⁻) to the weak acid (HN3). The equation is: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \( \text{pH} = 6.5 \) and \( \text{pKa} = -\log(1.9 \times 10^{-5}) \).

Calculate \( \text{pKa} \) using the given \( K_a \) value: \( \text{pKa} = -\log(1.9 \times 10^{-5}) \).

Rearrange the Henderson-Hasselbalch equation to solve for the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \): \( \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(\text{pH} - \text{pKa})} \).

Calculate the moles of HN3 in the solution using its concentration and volume: \( \text{moles of HN3} = 0.35 \text{ M} \times 1.8 \text{ L} \). Then, use the ratio from the Henderson-Hasselbalch equation to find the moles of NaN3 needed. Finally, convert moles of NaN3 to mass using its molar mass.