Which THE PH VALUE OF A solution 0,01 mol/l a monoacido that presents ka= 1×10-6
Well, the equation is
HA<<>>>H+ + A-
and ;etting x be the concentration of H+ and A- then
ka=x^2/(.01-x)
and put in Ka, and solve for x. It is a quadratic equaion.
Once you have x, then
pH=-log(x)
Thank you so much
To find the pH value of a solution with a concentration of 0.01 mol/L of a monoacidic solution with Ka = 1x10^(-6), you need to follow these steps:
Step 1: Write the balanced equation for the dissociation of the monoacidic solution. Let's assume the acid is represented as HA, and it dissociates into H+ and A-. The equation would be:
HA ⇌ H+ + A-
Step 2: Write the expression for the acid dissociation constant (Ka). In this case, Ka = [H+][A-]/[HA].
Step 3: Substitute the given value for Ka (1x10^(-6)) and the concentration of the solution (0.01 mol/L) into the Ka expression. The equation becomes:
1x10^(-6) = [H+][A-]/0.01
Step 4: Assuming complete dissociation of the acid (as it is a strong acid with a high Ka value), you can simplify the equation to:
1x10^(-6) = [H+][A-]
Step 5: Since the concentration of A- is equal to the concentration of H+ (assuming complete dissociation), you can substitute [H+] into the equation:
1x10^(-6) = ([H+])^2
Step 6: Solve for [H+] by taking the square root of both sides of the equation:
[H+] = sqrt(1x10^(-6))
Step 7: Calculate the hydrogen ion concentration ([H+]) using a calculator or by converting the scientific notation to a decimal:
[H+] = 1x10^(-3) mol/L or 0.001 mol/L
Step 8: Calculate the pH of the solution using the formula:
pH = -log[H+]
pH = -log(0.001)
Step 9: Evaluate the logarithm using a calculator:
pH ≈ 3
Therefore, the pH value of the solution with a concentration of 0.01 mol/L of a monoacidic solution with Ka = 1x10^(-6) is approximately 3.