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dlinn [17]
3 years ago
12

A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titra

ted with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H4O4 Mass of KHP used for standardization (g) 0.5053 Volume of NaOH required to neutralize KHP (mL) 13.4473 Calculate the concentration of NaOH solution in (mol/L)
Chemistry
1 answer:
Eddi Din [679]3 years ago
8 0

Answer:

See explanation below

Explanation:

In order to calculate this, we need to use the following expression to get the concentration of the base:

MaVa = MbVb (1)

We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:

Atomic weights of the elements to be used:

K = 39.0983 g/mol;  H = 1.0078 g/mol;  C = 12.0107 g/mol;  O = 15.999 g/mol

MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol

Now, let's calculate the mole of KHP:

moles = 0.5053 / 204.2189 = 0.00247 moles

With the moles, we also know that:

n = M*V (2)

Replacing in (1):

n = MbVb

Now, solving for Mb:

Mb = n/Vb  (3)

Finally, replacing the data:

Mb = 0.00247 / (13.4473/1000)

Mb = 0.184 M

This would be the concentration of NaOH

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Answer:

Here's what I get  

Explanation:

1. Write the chemical equation

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Let's rewrite the equation as

A⁻ + H₂O ⇌ HA + OH⁻

2. Calculate Kb

K_{\text{b}} = \dfrac{K_{\text{w}}}{K_{\text{a}}} = \dfrac{1.00 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10}

3. Set up an ICE table

                      A⁻ + H₂O ⇌ HA + OH⁻

I/mol·L⁻¹:      0.35                 0       0

C/mol·L⁻¹:       -x                  +x      +x

E/mol·L⁻¹:    0.35-x               x        x

4. Solve for x

\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}

Check for negligibility,

\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}

5. Calculate the pOH

[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹

pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88

6. Calculate the pH.

pH + pOH = 14.00

pH + 4.88 = 14.00

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Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.

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In this case, various reactants given will react exhibiting different types of chemical reactions to form products.

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