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nekit [7.7K]
3 years ago
14

What is acide, Base and salt​

Chemistry
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

In acid – base chemistry, salts are ionic compounds that result from the neutralization reaction of an acid and a base. Basic salts contain the conjugate base of a weak acid, so when they dissolve in water, they react with water to yield a solution with pH greater than 7.0.

Explanation:

Hope this helped!

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If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
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Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

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