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what is the ph of the final solution if 57.00 ml of 2.26 x 10-3m hcl is added to 52.00 ml of deionized water?

User Eltiare
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Answer:

pH = 3.93

Step-by-step explanation:

To calculate the pH of the final solution, we need to first calculate the moles of HCl that are added to the water.

moles of HCl = (volume in liters) x (molarity)

moles of HCl = (57.00 mL / 1000 mL/L) x (2.26 x 10^-3 mol/L)

moles of HCl = 1.29 x 10^-5 mol

Next, we need to calculate the total volume of the solution:

total volume = volume of HCl + volume of water

total volume = 57.00 mL + 52.00 mL = 109.00 mL = 0.109 L

Now we can use the moles of HCl and the total volume of the solution to calculate the molarity of the HCl in the final solution:

Molarity of HCl = moles of HCl / total volume of solution

Molarity of HCl = 1.29 x 10^-5 mol / 0.109 L

Molarity of HCl = 1.18 x 10^-4 M

Finally, we can use the molarity of the HCl to calculate the pH of the solution using the equation:

pH = -log[H+]

[H+] = molarity of HCl

pH = -log(1.18 x 10^-4)

pH = 3.93

Therefore, the pH of the final solution is approximately 3.93
User Hung Cao
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