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amm1812
2 years ago
6

Define ""acidity"" of an aqueous solution. How do you compare the strength of acidity of solutions ?

Engineering
1 answer:
Vlad [161]2 years ago
7 0

Answer with Explanation:

The acidity of an aqueous solution is a term used to identify how acidic the solution is. An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. In the other case around if  the concentration of hydrogen ions is lesser than the concentration of hydroxide ions the solution is termed to be basic or alkaline. For a solution with equal concentration of hydrogen and hydroxide ions the solution is termed to be neutral.

The acidity of solutions is compared on the basis of the concentration of the hydrogen ions reduced to log of base 10 to ease calculations. The comparison is made in terms of 'pH' value which is defined as

pH=-log[H^+]

where

 [H^+] is the hydrogen ion concentration of the solution in moles per liter of solution.

If the pH is < 7 the solution is acidic and the closer the pH value to 1 the higher is the acidity of the solution.

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Answer:

a) F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

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Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that X\sim Exp(\lambda)

And we know the probability denisty function for x given by:

f(x) = \lambda e^{-\lambda x} , x\geq 0

In order to find the cdf we need to do the following integral:

F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \

Part b

Assuming that X \sim Exp(\lambda =0.1), then the density function is given by:

f(x) = 0.1 e^{-0.1 x} dx , x\geq 0

And for this case we want this probability:

P(10 < X

And evaluating the integral we got:

P(10 < X

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