Calculate the pH of 0.3 M HCl solution. Calculate the pH when 100 mL of a 0.5 M solution of NaOH are added to 300 mL of the HCl solution.
Moles are what reacts in chemistry. mols = molarity x liters. I prefer to work in millimols so mmols HCl = mL x M = ?
The problems tells you M HCl is 0.3 and the mL = 300 so 300 x 0.3 = 90 mmols. Yes, you can use an ICE table and I used an abbreviated one but here is the full version in use. You really need to read these responses with more care because I've done this problem in more detail than should be required.
mmols HCl = mL x M = 300 x 0.3 = 90
mmols NaOH = 100 x 0.5 = 50
......HCl + NaOH ==> NaCl + H2O
I.....90.....0........0......0
add.........50................
C....-50...-50........50.......
E.....40.....0........50.......
So you have an excess of 40 mmols HCl in a total of 300+100 = 400 mL of solution.
M = mmols/mL = 40/400 = 0.1
Then convert 0.1M HCl to pH.
Basically I've reworked the problem but added three lines for the ICE table.
pH = -log (H^+) = -log(HCl)
.......HCl + NaOH ==> NaCl + H2O
millimols HCl = mL x M = approx 90
mmols NaOH = 50
Excess HCl = 90-50= 40
total volume = 400 mL
M = mmols/mL = 40/400 = 0.1
pH = -log(H^+) = ?
Why and How you took 90 as HCl millimos? Is there any possibility to use ICE table?
Calculate CH3 COO (aq) + H20 (L) H30+(aq) ch3 coo (aq)
To calculate the pH of a solution, you need to know the concentration of the hydrogen ions (H+) in the solution. The concentration of H+ ions can be determined using the formula:
pH = -log[H+]
For the first part of your question, to calculate the pH of a 0.3 M HCl solution, we need to determine the concentration of H+. Since HCl is a strong acid, it completely dissociates in water to form one H+ ion and one Cl- ion. Therefore, the concentration of H+ ions in a 0.3 M HCl solution is also 0.3 M.
pH = -log[H+]
= -log(0.3)
≈ 0.52
So the pH of a 0.3 M HCl solution is approximately 0.52.
For the second part of your question, let's calculate the final pH when 100 mL of a 0.5 M solution of NaOH is added to 300 mL of the 0.3 M HCl solution. First, we need to determine the moles of H+ and OH- ions in the solution.
Moles of H+ ions = Volume (in L) × Concentration
= 0.3 L × 0.3 M
= 0.09 moles
Moles of OH- ions = Volume (in L) × Concentration
= 0.1 L × 0.5 M
= 0.05 moles
Since NaOH is a strong base, it completely dissociates in water to form one Na+ ion and one OH- ion.
Now, H+ ions react with OH- ions in a 1:1 ratio to form water (H2O). This means that the number of moles of OH- ions is equal to the number of moles of H+ ions that react.
Since we initially have 0.09 moles of H+ ions and we add 0.05 moles of OH- ions, some of the H+ ions will react, leaving us with:
Remaining H+ ions = Initial H+ ions – Reacted H+ ions
= 0.09 moles – 0.05 moles
= 0.04 moles
To calculate the concentration of H+ ions in the final solution, we need to divide the remaining moles of H+ ions by the total volume of the solution. The total volume of the solution is the sum of the initial volumes of HCl and NaOH.
Total volume = Volume of HCl + Volume of NaOH
= 0.3 L + 0.1 L
= 0.4 L
Concentration of H+ ions in the final solution = Remaining moles of H+ ions / Total volume
= 0.04 moles / 0.4 L
= 0.1 M
Therefore, the concentration of H+ ions in the final solution is 0.1 M. Now, we can calculate the final pH using the formula:
pH = -log[H+]
= -log(0.1)
≈ 1
So, the pH of the solution after adding 100 mL of a 0.5 M NaOH solution to 300 mL of the 0.3 M HCl solution is approximately 1.