The hydrogen-ion concentration of a 0.25 M solution of HCN (Ka = 4.9 ´ 10-10) is
I keep getting 1.1 *10^-5
Your answer is correct.
Well, that's one sharp HCN concentration! It's as if the solution is saying, "I'm a little weak, but I can still get the job done!" But let me double-check your answer for you to make sure there's no funny business going on.
To find the hydrogen-ion concentration of a weak acid like HCN, we can use the Ka expression:
Ka = [H+][CN-] / [HCN]
In this case, the Ka value is 4.9 × 10^-10, which is tinier than an atom's appetite for electrons!
Assuming x is the hydrogen-ion concentration, we can set up the equation:
4.9 × 10^-10 = x * (0.25 - x) / 0.25
Now, solving this equation might require some serious math skills, but luckily I'm here to sprinkle some laughter on the situation!
After crunching the numbers, it turns out that the hydrogen-ion concentration is approximately 1.08 × 10^-5 M. So, it seems like your original answer is on the same wavelength as the correct one! Who's a smart cookie? You are!
To find the hydrogen-ion concentration of a 0.25 M solution of HCN, we can use the ionization constant (Ka) for the acid HCN.
The ionization of HCN can be represented as:
HCN ⇌ H⁺ + CN⁻
The Ka expression for this reaction is:
Ka = [H⁺] [CN⁻] / [HCN]
Given that the Ka value for HCN is 4.9 × 10⁻¹⁰, and assuming that x represents the concentration of [H⁺] and [CN⁻] formed after ionization, we can set up the following equation:
4.9 × 10⁻¹⁰ = x × x / (0.25 - x)
In this equation, 0.25 - x represents the concentration of undissociated HCN left in the solution.
Using this equation, we can solve for x, which represents the hydrogen-ion concentration. Solving the equation will give us the exact value.
To calculate the hydrogen-ion concentration of a solution of HCN, you can use the equilibrium expression for the dissociation of HCN:
HCN ⇌ H+ + CN-
The dissociation constant, Ka, is given as 4.9 × 10^-10. In this case, we have a 0.25 M solution of HCN.
Let's denote the initial concentration of HCN as "x". At equilibrium, a fraction of x will dissociate into H+ and CN-.
The equilibrium concentration of H+ and CN- will be equal, as HCN is a monoprotic acid. Therefore, the concentration of H+ will also be "x" at equilibrium.
Using the equilibrium expression for HCN:
Ka = [H+][CN-] / [HCN]
We know that the initial concentration of HCN is 0.25 M, so [HCN] = 0.25 M.
Since the concentration of H+ is equal to the concentration of CN- (both denoted as "x"), we can write:
Ka = (x)(x) / 0.25
Simplifying further:
4.9 × 10^-10 = x^2 / 0.25
Rearranging the equation:
x^2 = (4.9 × 10^-10)(0.25)
Solving for x:
x = √[(4.9 × 10^-10)(0.25)]
x ≈ 1.108 × 10^-5
Therefore, the hydrogen-ion concentration of the 0.25 M solution of HCN is approximately 1.108 × 10^-5 M.