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To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the resulting solution is diluted to 100 ml. What is change in pH of the HCl solution?
- a)4.398
- b)0.398
- c)0.1 M
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the...
**Explanation:**
To find the change in pH of the HCl solution, we need to determine the concentration of H+ ions before and after the addition of NaOH.
**Step 1: Determine the initial concentration of H+ ions in the HCl solution**
Given:
Volume of HCl solution = 50 ml
Molarity of HCl solution = 0.1 M
Using the formula:
Molarity (M) = moles of solute / volume of solution (L)
We can rearrange the formula to find the moles of solute:
moles of solute = Molarity × volume of solution (L)
Converting the volume of HCl solution to liters:
Volume of HCl solution = 50 ml = 50/1000 L = 0.05 L
Moles of HCl = 0.1 M × 0.05 L = 0.005 moles
Since HCl is a strong acid, it completely dissociates in water to form H+ ions. Therefore, the initial concentration of H+ ions in the HCl solution is also 0.1 M.
**Step 2: Determine the concentration of H+ ions after the addition of NaOH**
Given:
Volume of NaOH solution = 10 ml
Molarity of NaOH solution = 0.1 M
Converting the volume of NaOH solution to liters:
Volume of NaOH solution = 10 ml = 10/1000 L = 0.01 L
Moles of NaOH = 0.1 M × 0.01 L = 0.001 moles
Since NaOH is a strong base, it completely dissociates in water to form OH- ions. Therefore, the concentration of OH- ions added to the solution is also 0.1 M.
**Step 3: Determine the new concentration of H+ ions after neutralization**
Since HCl and NaOH react in a 1:1 ratio, the moles of H+ ions neutralized by OH- ions is equal to the moles of NaOH added.
Therefore, the concentration of H+ ions after neutralization = initial concentration of H+ ions - concentration of H+ ions neutralized
= 0.1 M - 0.001 M = 0.099 M
**Step 4: Determine the change in pH**
pH is defined as the negative logarithm of the concentration of H+ ions.
Initial pH = -log(0.1) = 1
New pH = -log(0.099) = 1.398
Change in pH = New pH - Initial pH = 1.398 - 1 = 0.398
Therefore, the change in pH of the HCl solution is 0.398, which corresponds to option B.
To find the change in pH of the HCl solution, we need to determine the concentration of H+ ions before and after the addition of NaOH.
**Step 1: Determine the initial concentration of H+ ions in the HCl solution**
Given:
Volume of HCl solution = 50 ml
Molarity of HCl solution = 0.1 M
Using the formula:
Molarity (M) = moles of solute / volume of solution (L)
We can rearrange the formula to find the moles of solute:
moles of solute = Molarity × volume of solution (L)
Converting the volume of HCl solution to liters:
Volume of HCl solution = 50 ml = 50/1000 L = 0.05 L
Moles of HCl = 0.1 M × 0.05 L = 0.005 moles
Since HCl is a strong acid, it completely dissociates in water to form H+ ions. Therefore, the initial concentration of H+ ions in the HCl solution is also 0.1 M.
**Step 2: Determine the concentration of H+ ions after the addition of NaOH**
Given:
Volume of NaOH solution = 10 ml
Molarity of NaOH solution = 0.1 M
Converting the volume of NaOH solution to liters:
Volume of NaOH solution = 10 ml = 10/1000 L = 0.01 L
Moles of NaOH = 0.1 M × 0.01 L = 0.001 moles
Since NaOH is a strong base, it completely dissociates in water to form OH- ions. Therefore, the concentration of OH- ions added to the solution is also 0.1 M.
**Step 3: Determine the new concentration of H+ ions after neutralization**
Since HCl and NaOH react in a 1:1 ratio, the moles of H+ ions neutralized by OH- ions is equal to the moles of NaOH added.
Therefore, the concentration of H+ ions after neutralization = initial concentration of H+ ions - concentration of H+ ions neutralized
= 0.1 M - 0.001 M = 0.099 M
**Step 4: Determine the change in pH**
pH is defined as the negative logarithm of the concentration of H+ ions.
Initial pH = -log(0.1) = 1
New pH = -log(0.099) = 1.398
Change in pH = New pH - Initial pH = 1.398 - 1 = 0.398
Therefore, the change in pH of the HCl solution is 0.398, which corresponds to option B.
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Community Answer
To a 50 ml of 0.1 M HCl solution, 10 ml of 0.1 M NaOH is added and the...
Before pH=1
after reaction with dilute
concentration of H+ =4/100= 0.04 pH=1.398
difference∆Ph =o.398
after reaction with dilute
concentration of H+ =4/100= 0.04 pH=1.398
difference∆Ph =o.398
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