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Tomtit [17]
3 years ago
9

How do scientists prevent the nonpolar perfluorocarbons

Chemistry
1 answer:
Pavel [41]3 years ago
3 0

Scientists prevent the nonpolar perfluorocarbons

in Oxygent from separating from the water by adding surfactant.

<u>Explanation:</u>

The  collection of oxygen from lungs and transporting it  all those tissues present in the human body and taking carbon dioxide back to lungs are the main functions of the hemoglobin. But, instead of this hemoglobin, When perfluorocarbons , a nonpolar chemical is used to perform the same function for the oxygen transportation, the nonpolar chemical may not mix well with the water when this perfluorocarbons is taken through a water-based saline .

For this purpose an addition of a chemical called surfactant is done. This will enhance the chemical bonding between the non polar chemical and water. The perfluorocarbons  spreads into some droplets and gets coated with this bonding substance. Thus it gets attached to the water.

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Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans
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The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 10.4

<u>Explanation:</u>

We are given:

Molarity of sodium hypochlorite = 0.39 M

pK_a of HClO = 7.50

We know that:

pK_a=-\log K_a

K_a  of HClO = 10^{-7.50}=3.16\times 10^{-8}

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant  = 3.16\times 10^{-8}

K_b = Base dissociation constant

Putting values in above equation, we get:

10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}

The chemical equation for the reaction of hypochlorite ion with water follows:

                    ClO^-+H_2O\rightarrow HClO+OH^-

<u>Initial:</u>           0.39

<u>At eqllm:</u>      0.39-x                   x           x

The expression of K_b for above equation follows:

K_b=\frac{[HClO][OH^-]}{[ClO^-]}

Putting values in above equation, we get:

3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.00035M

Putting values in above equation, we get:

pOH=-\log (0.00035)=3.6

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-3.6=10.4

Hence, the pH of the solution is 10.4

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