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3 years ago

fish give off the compound ammonia which has a pH above 7 to which class of compounds does ammonia belong

Chemistry

1 answer:

jasenka [17]3 years ago

5 0

Answer:
The compound ammonia given by fish is alkaline

Explanation:
We can classify elements/compounds based on their pH values into three types:
acids: these are compounds having pH value lower than 7
neutral: these are compounds having pH value equal to 7
alkalies: these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)

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3 years ago

1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o

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Answer:

* $pH=11.0$

* $[HClO]_{eq}=9.32x10^{-4}M$

* $[HF]_{eq}=5.07x10^{-6}M$

Explanation:

Hello,

At first, the molarities of NaClO and KF are computed as shown below:

$M_{NaClO}=\frac{51.2g*\frac{1mol}{74.45g} }{0.250L}=2.75M \\\\M_{KF}=\frac{25.3g*\frac{1mol}{58.1g} }{0.250L} =1.74M$

Now, since both NaClO and KF are ionic, one proposes the dissociation reactions as:

$NaClO\rightarrow Na^++ClO^-\\KF\rightarrow K^++F^-$

In such a way, by writing the formation of HClO and HF respectively, due to the action of water, we obtain:

$ClO^-+H_2O\rightleftharpoons HClO+OH^-\\F^-+H_2O\rightleftharpoons HF+OH^-\\$

Now, the presence of OH⁻ immediately implies that Kb for both NaClO and KF must be known as follows:

$Kb=\frac{Kw}{Ka} \\Kb_{NaClO}=\frac{1x10^{-14}}{10^{-7.50}}=3.16x10^{-7}\\Kb_{KF}=\frac{1x10^{-14}}{10^{-3.17}}=1.48x10^{-11}$

In such a way, the law of mass action for each case is:

$Kb_{NaClO}=3.16x10^{-7}=\frac{[HClO][OH]^-}{[ClO^-]}=\frac{(x_{ClO})(x_{ClO})}{2.75-x_{ClO}}\\Kb_{KF}=1.48x10^{-11}=\frac{[HF][OH]^-}{[F^-]}=\frac{(x_{F})(x_{F})}{1.74-x_{F}}$