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NeX [460]
2 years ago
8

If you have 250mL of an 11M solution and add enough water to dilute it to 7 M, then what is the final volume of the solution?

Chemistry
1 answer:
Luba_88 [7]2 years ago
4 0
35.7M is the answer to this question
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B

Explanation:

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What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso
Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

 

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M 
[A-]= 0.609/ 2.00 = 0.3045 M 
pKa = 6.25 

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0
3 years ago
Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0
3 years ago
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