whats the pH if 25.0 ml of 2.0 M NaOH is added to 1.00 L of pure water?
First calculate the molarity of the NaOH. It is diluted 25 mL to 1,025 mL. So 2.0 M x (25/1025) = ??
NaOH is a strong electrolyte and is ionized 100%; therefore, OH^- = the molarity. Take it from here with the following:
pOH = - log (OH^-)
pH + pOH = 14.
Pure water is added to 25.0 mL of a 1.00 M HCL solution. The new solution has a volume of 2.00 L. What is the pH of the new solution.
To find the pH of the solution, we need to calculate the concentration of hydroxide ions (OH-) added by the NaOH and use that to determine the pOH. From there, we can calculate the pH using the equation:
pH = 14 - pOH
First, we need to find the number of moles of NaOH added:
moles of NaOH = volume of NaOH (L) x concentration of NaOH (M)
= 0.025 L x 2.0 M
= 0.05 mol
Since NaOH is a strong base, it fully dissociates in water to produce hydroxide ions (OH-). Therefore, the number of moles of hydroxide ions added is the same as the number of moles of NaOH:
moles of OH- = 0.05 mol
Next, we calculate the concentration of hydroxide ions (OH-):
concentration of OH- = moles of OH- / volume of water (L)
= 0.05 mol / 1.00 L
= 0.05 M
Now, we can calculate the pOH by taking the negative logarithm of the hydroxide ion concentration:
pOH = -log10(0.05)
= -(-1.301)
= 1.301
Finally, we can calculate the pH using the equation pH = 14 - pOH:
pH = 14 - 1.301
= 12.699
Therefore, the pH of the solution after adding 25.0 mL of 2.0 M NaOH to 1.00 L of pure water is approximately 12.699.
To determine the pH of a solution, we first need to calculate the concentration of hydroxide ions (OH-) in the solution. Since NaOH is a strong base that dissociates completely in water, we can use stoichiometry to find the concentration:
1. Calculate the number of moles of NaOH:
Moles of NaOH = concentration of NaOH × volume of NaOH
= 2.0 M × 0.025 L
= 0.05 moles
2. Calculate the volume of the solution:
Volume of the solution = volume of pure water + volume of NaOH
= 1.00 L + 0.025 L
= 1.025 L
3. Calculate the concentration of hydroxide ions:
Concentration of OH- = moles of OH- / volume of the solution
= 0.05 moles / 1.025 L
≈ 0.049 M
Now, to find the pOH (negative logarithm of hydroxide ion concentration), we can use the formula:
pOH = -log10(OH-) (logarithm to the base 10)
4. Calculate the pOH:
pOH = -log10(0.049)
≈ 1.31
Finally, to find the pH, we can use the relationship:
pH + pOH = 14
5. Calculate the pH:
pH = 14 - pOH
= 14 - 1.31
≈ 12.69
Therefore, the pH of the solution, when 25.0 mL of 2.0 M NaOH is added to 1.00 L of pure water, is approximately 12.69.