1. Find the pH when there are 0.034 moles of h+ in a liter of solution
2. Find the pH when there are 1.6x10^-7 moles of h+ in 500 ml of solution
Please help me on these only problems im struggling with, thank you!
1. Well, if we have 0.034 moles of H+ in a liter of solution, it means we have a pretty acidic situation. But let's find the pH using a little chemistry humor: Why did the proton bring a calculator to the party? Because it wanted to find the pH-ion!
To find the pH, we need to use the formula: pH = -log[H+]. Since we have 0.034 moles of H+, the concentration [H+] is 0.034 M (moles per liter). Taking the negative logarithm of 0.034, we find that the pH is approximately 1.468. So, the pH-ion... I mean pH value in this case is 1.468.
2. Now, in this case, we have 1.6x10^-7 moles of H+ in 500 ml of solution. Let's get pH-tastic with it! How do you recognize a happy molecule? It has a pHace!
To find the pH, we need to calculate the concentration of H+ first. Since we have 1.6x10^-7 moles of H+ in 500 ml (which is 0.5 liters) of solution, the concentration [H+] is (1.6x10^-7 moles) / (0.5 liters) = 3.2x10^-7 M (moles per liter).
Using the pH formula: pH = -log[H+], we can find that the pH is approximately 6.495. So, the pHace... I mean pH value in this case is 6.495.
I hope these pun-filled explanations brightened up your chemistry problems a little!
Sure! I can help you with these problems.
1. To find the pH when there are 0.034 moles of H+ in a liter of solution, we can use the formula for pH:
pH = -log[H+]
In this case, [H+] refers to the concentration of H+ ions in moles per liter. So in the given solution, [H+] = 0.034 moles/L.
Using the formula, we can calculate the pH as follows:
pH = -log(0.034)
= -(-1.47)
= 1.47
Therefore, the pH of the solution is approximately 1.47.
2. To find the pH when there are 1.6x10^-7 moles of H+ in 500 ml of solution, we need to convert the volume to liters and then use the same formula as above.
Given that the volume is 500 ml, we first convert it to liters by dividing by 1000:
Volume = 500 ml / 1000
= 0.5 L
Now, we can use the formula to calculate the pH:
pH = -log[H+]
[H+] = 1.6x10^-7 moles / 0.5 L
= 3.2x10^-7 moles/L
pH = -log(3.2x10^-7)
= -(-6.5)
= 6.5
Therefore, the pH of the solution is approximately 6.5.
I hope this helps! Let me know if you have any further questions.
Sure! I'll help you with both problems step by step.
1. Find the pH when there are 0.034 moles of H+ in a liter of solution:
To find the pH, we need to use the formula for pH, which is defined as the negative logarithm (base 10) of the concentration of H+ ions in the solution.
Step 1: Calculate the concentration of H+ ions.
The concentration is given as 0.034 moles per liter. This means there are 0.034 moles of H+ ions in 1 liter of solution.
Step 2: Take the negative logarithm.
To find the pH, take the negative logarithm of the concentration of H+ ions. Mathematically, this can be written as pH = -log[H+].
pH = -log(0.034)
Step 3: Calculate the pH.
Using a calculator, enter -log(0.034) and compute the value. The result will be the pH of the solution.
2. Find the pH when there are 1.6x10^-7 moles of H+ in 500 ml of solution:
Step 1: Convert the volume to liters.
Since the concentration is given per liter, we need to convert the volume to liters. There are 1000 milliliters (mL) in one liter. Therefore, 500 mL is equivalent to 0.5 liters.
Step 2: Calculate the concentration of H+ ions.
The concentration is given as 1.6x10^-7 moles per 0.5 liters. This means there are 1.6x10^-7 moles of H+ ions in 0.5 liters of solution.
Step 3: Take the negative logarithm.
Following the same formula as before, we take the negative logarithm of the concentration: pH = -log[H+].
pH = -log(1.6x10^-7)
Step 4: Calculate the pH.
Using a calculator, enter -log(1.6x10^-7) and compute the value. The result will be the pH of the solution.
I hope that helps! Let me know if you have any further questions.
It would help if you told us what you don't understand about these.
For #1, pH = -log(H^+). Substitute and solve for pH.
For #2. Same formula. I assume this is the TOTAL H^+ and there is no need to add in any contribution due to the ionization of H2O.