Calculate [H3O+] at 25 ∘C for each solution and determine if the solution is acidic, basic, or neutral.a.) [OH−] = 3.8×10−2 Mb.) [OH−] = 1.0×10-7 Mc.) [OH−] = 5.5×10−10 MPlease show the work and how you determine if the solution is neutral, acidic, or basic

The concentration of H⁺ ion's (pH) of a solution is the magnitude of negative sign of logarithm value of [H₃O⁺}, is called pH of that solution.

That is

[tex]pH=-log[H_3O^+][/tex]

We know that,

[tex][H_3O^+][OH^-]=K_w=10^{-14}[/tex] [ at 25°C]

Taking log both sides

[tex]log[H_3O^+]+log[OH^-]=logK_w=log 10^{-14}[/tex]

[tex]\Rightarrow - log[H_3O^+]-log[OH^-]=-logK_w=-log 10^{-14}[/tex]

[tex]\Rightarrow pH+pOH=pK_w={14}[/tex]

[tex]\therefore pH+pOH={14}[/tex]

(a)

[tex][OH^-]=3.8\times10^{-2}M[/tex]

[tex]- log[H_3O^+]-log[OH^-]=14}[/tex]

[tex]\Rightarrow pH-log(3.8\times 10^{-2})=14[/tex]

[tex]\Rightarrow pH-(-1.42)=14[/tex]

[tex]\Rightarrow pH=14-1.42[/tex]

∴pH= 12.58

since pH>7

The solution is basic in nature.

(b)

[tex][OH^-]=1.0\times10^{-7}M[/tex]

[tex]- log[H_3O^+]-log[OH^-]=14}[/tex]

[tex]\Rightarrow pH-log(1.0\times 10^{-7})=14[/tex]

[tex]\Rightarrow pH-(-7)=14[/tex]

[tex]\Rightarrow pH=14-7[/tex]

∴pH= 7

The solution is neutral in nature.

(c)

[tex][OH^-]=5.5\times10^{-10}M[/tex]

[tex]- log[H_3O^+]-log[OH^-]=14}[/tex]

[tex]\Rightarrow pH-log(5.5\times 10^{-10})=14[/tex]

[tex]\Rightarrow pH-(-9.26)=14[/tex]

[tex]\Rightarrow pH+9.26=14[/tex]

[tex]\Rightarrow pH=14-9.26[/tex]

∴pH= 4.74

Since pH<7

The solution is acidic in nature.