og IOH thus giving the By defini ion, pH og (H,0 and poH the above equation. rel

Question

og IOH thus giving the By defini ion, pH og (H,0 and poH the above equation. relationship of pH poH 14.00 at 25 C from B. Acid Solution is a species (molecule or ion) According to the Bronsted-Lowry theory, an acid (molecule or ion) that accepts that donates a proton (H) and a base is a species a proton In aqueous solution, the substance HA an acid because it donates a proton to s the water molecule, which acts as a base in accepting it, ie, A quantitative measure of the extent of ionization of an acid is given by its equi librium constant, K. [HA] (Remember, water does not appear in the K expression as the concentration of water is essentially constant in aqueous solutions. The ]brackets refer to molar concentrations.) For strong acids, K. 1, and the HA is essentially 100% ionized. Thus, for strong acids at equilibrium, we know [HA] 0 and [H,0 [A l Thus, you can calculate the [H,0 from the initial molar concentration of the strong acid and likewise the pH of the resulting solution, log [H,0 pH log [HA] For weak acids, K, 1, and the HA is only partially ionized in solution. Thus, for weak acids at equilibrium, we know: IA and IHA [HA] When K 1, the extent of ionization is very small and the assumption can often be made that [HA Al iur In this experiment, we can safely make that assumption. By measuring the pH of a weak acid at equilibrium, we can calculate the [H,0 l and hence the K, value if we know the initial HA molar concentration.

Explanation / Answer

A. pH of acid solutions

This is the information we have:

Now, to get the rest of the information, let's start with the molarity of the second solution HCl (2).

According to the procedure, we diluted 1mL (V1) of HCl 0.1M (C1) to 10mL (V2) to have the HCl (2). So, this means HCl (2) is ten times more diluted than solution one. Let's use the CV formula to make sure of the concentration:

C1V1 = C2V2

0.1M(1mL) = C2(10mL)

C2 = 0.01M

Which is ten times more diluted than C1.

So our table now looks like this:

Now, for the calculated pH we need to use the formula:

pH = -log(H+)

Since we have a strong acid here, the molarity is equal to H+.

So for the HCl (1) solution:

pH = -log(0.1)

pH = 1

And for the HCl (2) solution:

pH = -log(0.01)

pH = 2

Strong acid HCl (1) HCl (2) Measured pH 1.37 2.14 Molarity 0.1M Calculated pH

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