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NikAS [45]
3 years ago
10

You are going to standardize your sodium hydroxide by titrating with potassium hydrogen phthalate. As an example, you dissolve 0

.3365 g of potassium hydrogen phthalate, otherwise known as KHP (KHC8H4O4) in water in a 250.0 mL Erlenmeyer flask and then add phenolphthalein indicator. You then titrate with your sodium hydroxide solution, which is in a buret, and you determine that the equivalence point is at 12.44 mL of your sodium hydroxide solution. Determine the molarity of your sodium hydroxide solution. What color will the solution of potassium hydrogen phthalate turn to determine when the equivalence point has been reached
Chemistry
1 answer:
Kay [80]3 years ago
4 0

Answer:

0.13 M

Explanation:

The reaction equation is;

NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)

Molar mass of KHP = 204.22 g/mol

Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles

n= CV

Where;

C= concentration

V= volume in dm^3

n= number of moles

C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M

If 1 mole of KHP reacts with 1 mole of NaOH

1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH

From

n= CV

We have that only 12.44 ml of NaOH reacted

C= n/V = 1.65 × 10^-3 moles × 1000/12.44

C= 0.13 M

At the equivalence point, the KHP solution turned light pink.

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Read 2 more answers
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

3 0
3 years ago
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