Just checking to see if my answers are correct.
5mL of 0.1M NaOH and 5mL water.
Pouring 1mL of solution from 1st test tube, 2nd to 3rd, etc.
Concentration NaOH .1M
Concentration pOH pH
#1) .05 -log(.05)=1.3 14-1.3=12.7
#2) .005 -log(.005)=2.3 14-2.3=11.7
hi iwant to answer question
ok. There should be spaces between 0.05M pOH= -log(.05)=1.3 pH= 14-2.3=11.7
TT #2
Concentration =0.005M
pOH= -log(0.005)=2.3
pH= 14-2.3=11.7
Your calculations seem accurate. Here's a step-by-step explanation of how you arrived at the answers:
1) To find the pOH, you first need to calculate the concentration of hydroxide ions (OH-) in the solution. In the first test tube, you have 5mL of 0.1M NaOH, which means you have 5mmol of NaOH. Since NaOH dissociates completely in water, you have 5mmol of OH- ions.
To convert from moles to concentration, you divide the moles of OH- by the total volume in liters. In this case, the total volume is 10mL (5mL NaOH + 5mL water) or 0.01L. So the concentration of OH- is 5mmol/0.01L = 0.5M.
Now, you can calculate the pOH by taking the negative logarithm (base 10) of the OH- concentration. pOH = -log(0.5) = 1.3.
To find the pH, you use the fact that pH + pOH = 14. So, pH = 14 - pOH = 14 - 1.3 = 12.7.
2) Following the same process as before, you have 0.005M concentration of OH-. The pOH is then -log(0.005) = 2.3. The pH is calculated as 14 - pOH, so pH = 14 - 2.3 = 11.7.
Based on your calculations, it seems like you've correctly determined the pOH and pH values for the given solutions. Well done!