What is the pH of the solution created by combining 2.50 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
I believe if my calculations are correct the answer is 1.28
however i do not know how to answer this question because it has a weak acid instead of a strong acid
What is the pH of the solution created by combining 2.50 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)?
Furthermore, I have no idea what to do for this question...
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water
hint: for the strong base& weak acid
steps
1. write the balanced eq of the acid and base
2. determine if moles H+ or OH- is in excess
3. determine amount of excess
4. determine Molarity of excess amount
5. solve for pH directly or by pOH if OH- is in excess
For the dilution: pH isn't affected by dilution unless it's in an extremely large ammount of water (I want to say a million liters but let's just say alot, I can't remember) In general it doesn't affect pH
Your first answer is correct
To find the pH of a solution created by combining a strong base (NaOH) with a strong acid (HCl), you can use the following steps:
1. Write the balanced equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
2. Determine if moles of H+ or OH- are in excess. In this case, since NaOH is a strong base and HCl is a strong acid, the reaction goes to completion and all of the NaOH reacts with HCl. Therefore, there are no excess moles of H+ or OH-.
3. Determine the amount of excess. Since there is no excess, this step is not applicable.
4. Determine the molarity of the excess amount. Since there is no excess, this step is not applicable.
5. Solve for pH directly or by pOH if OH- is in excess. In this case, since there is no excess OH-, we can calculate the pH directly.
To do so, we can use the equation for the reaction between HCl and NaOH to determine the moles of HCl that react with 2.50 mL of NaOH:
0.10 moles/L * 0.0025 L = 0.00025 moles of HCl
Since the reaction is 1:1, the number of moles of H+ produced is equal to the number of moles of HCl used.
So, we have 0.00025 moles of H+ in a total volume of 10.50 mL (2.50 mL + 8.00 mL). To find the molarity of H+ ions, we divide the moles of H+ by the total volume in liters:
0.00025 moles / 0.0105 L = 0.0238 M
Finally, to find the pH, we use the equation: pH = -log[H+]
pH = -log(0.0238) = 1.62
Therefore, the pH of the solution created by combining 2.50 mL of 0.10 M NaOH with 8.00 mL of 0.10 M HCl is approximately 1.62.
(Since you mentioned the answer as 1.28, please double-check your calculations to ensure accuracy.)
For the second question involving a weak acid, HC2H3O2 (acetic acid), the steps would be slightly different. Acetic acid is a weak acid, so the reaction doesn't go to completion. You would need to consider the dissociation constant (Ka) of acetic acid to determine the concentration of H+ ions in solution.
If you provide me with the Ka value for acetic acid, I can guide you through the process of finding the pH.