Determine the pH?

Question

A 200.0 mL sample of 0.10 M Sr(OH)₂ is titrated with 0.10 M HCl. Determine the pH of the solution after the addition of 800.0 mL HCl.

Answer 1

First, remember that HCl is a strong acid that will dissociate completely into H⁺ and Cl^-. So we write the equation for the dissociation of HCl like this: HCl --> H⁺ + Cl^-. Also remember that Sr(OH)2 is a strong base that will dissociate completely into Sr²⁺ and 2 OH^-. So we write the equation for the dissociation of Sr(OH)2 like this: Sr(OH)₂ --> Sr²⁺ + 2 OH^-. And finally, we can write the overall equation for the overall reaction between HCl and Sr(OH)2 like this: 2 HCl + Sr(OH)₂ --> SrCl₂ + 2 H₂O

Now we need to find the moles of HCl. The problem tells us that we have 800 mL of 0.10 M HCl. We convert 800 mL to L with the conversion factor 1000 mL = 1 L. So 800 mL = 0.800 L. Now we find the moles of HCL by multiplying 0.800 L x 0.10 M. 0.800 x 0.10 = 0.08 moles of HCl. So before the HCl is added to the Sr(OH)₂, we have 0.08 moles of HCl.

Now we find the moles of Sr(OH)₂. The problem tells us that we have 200 mL of 0.10 Sr(OH)₂. Remember that 1000 mL = 1 L, so 200 mL = 0.2 L. We find the moles of Sr(OH)₂ by multiplying 0.200 L x 0.10 M.

0.200 x 0.10 = 0.02. So we have 0.02 moles of Sr(OH)₂ before the Sr(OH)₂ is added to the HCl.

How many moles of HCl will be used up when neutralizing the 0.02 moles of Sr(OH)₂? If you look back at the overall equation, it shows that 2 moles of HCl are needed to neutralize each mole of Sr(OH)₂. So, we are going to use up twice the moles of Sr(OH)₂. We already found out that we have 0.02 moles of Sr(OH₎₂, so we are going to use up 2 x 0.02 = 0.04 moles of HCL to neutralize the Sr(OH)₂. So if we started with 0.08 moles of HCl, and we use 0.04 moles of HCl to neutralize the Sr(OH)₂, we have 0.08 -- 0.04 = 0.04 moles of HCl left over. (We don't have any Sr(OH)₂ left over because we neutralized it.) Now we have to calculate the concentration (the M) for the HCl that is left. We know we have 0.04 moles of HCl left, but the volume has changed. The volume is now 0.2 mL + 0.8 mL = 1 L. So the concentration of HCl is 0.04 mol / 1 L = 0.04 M HCl. This also means that the concentration of H⁺ is 0.04.

Now we find the pH by doing --log (0.04). We use a calculator to figure out that --log (0.04) = 1.3979, which we round to 1.4. So the pH is 1.4.

Answer 2

Step 1: Calculate the number of moles of OH^-

Step 2: Calculate the number of moles of H⁺

Step 3: Determine which is excess and how much is leftover

Step 4: Divide by total volume to calculate concentration

Step 5: Calculate pH

Step 1: To calculate moles from molarity, we have to multiply by volume.

0.10 M x 0.2000 L = 0.020 moles of Sr(OH)₂.

Note that every 1 mole of Sr(OH)₂ will dissociate into 2 OH^-. Thus, the number of moles of OH^- will be twice the moles of Sr(OH)₂ = 0.040 moles

Step 2: 0.10 M x 0.8000 L = 0.080 moles HCl = 0.080 moles of H⁺.

Step 3: We have more moles of acid (H⁺) than base (OH^-). Thus the solution will be acidic. H⁺ and OH^- react in a 1-to-1 ratio to form water according to the formula:

H⁺ + OH^- -> H₂O

Since they react in a 1:1 ratio, the excess can be calculated by simply subtracting: 0.080 - 0.040 = 0.040 moles H⁺

Step 4: To calculate pH, we need to figure out the concentration of [H⁺]. Right now, we know moles. To convert to molarity, we must divide by the total volume. If we mixed a 200.0 mL solution with 800.0 mL of another solution, the total volume of the resultant solution is 1000.0 mL or 1.0000 L.

M = n/V = 0.000 moles / 1.0000 L = 0.040 M

Step 5: pH = -Log[H⁺]

pH = -Log(0.040)

pH = 1.40 (2 sig figs)