Solveeit Logo
Login

Question

Question: An alcoholic drink substance pH = 4.7 then \(O{{H}^{-}}\)concentration of this solution is:- (\({{...

An alcoholic drink substance pH = 4.7 then OH−O{{H}^{-}}concentration of this solution is:-
(Kw=10−14mol2L−2{{K}_{w}}={{10}^{-14}}mo{{l}^{2}}{{L}^{-2}})
(A) 3×10−103\times {{10}^{-10}}
(B) 5×10−105\times {{10}^{-10}}
(C) 1×10−101\times {{10}^{-10}}
(D) 5×10−85\times {{10}^{-8}}

Explanation

Solution

As we know that pH is the abbreviation for power of hydrogen and it tells about the concentration of hydrogen ion (H+{{H}^{+}}) in a solution whereas pOH is the abbreviation for power of hydroxide ion and it tells about the concentration of hydroxide ion (OH−O{{H}^{-}} ) in a solution. So here we have to calculate the concentration of hydroxide ion in the given substance.
Formula used:
We will use the following formulas in this solution:-
pH + pOH = 14
pOH = −log⁡[OH−]-\log [O{{H}^{-}}]

Complete answer:
Let us first discuss about pH and pOH followed by further calculations:-
-pH: It is also known as the power of hydrogen and it informs regarding the concentration of hydrogen ions (H+{{H}^{+}}) in a solution. It is calculated by negative logarithm of concentration of hydrogen ions i.e., pH = −log⁡[H+]-\log [{{H}^{+}}] .
-pOH: It is also known as power of hydroxide ion and it informs regarding the concentration of hydroxide ion (OH−O{{H}^{-}}) in a solution. It is calculated by negative logarithm of concentration of hydroxide ion i.e., pOH = −log⁡[OH−]-\log [O{{H}^{-}}] .
The relation between pH and pOH of any substance or solution is pH + pOH = 14.
-Calculation of pOH of alcoholic drink substance using the relation between pH and pOH as follows:-
The given value of pH of alcoholic drink substance is = 4.7
On substituting this value in the formula, we get:-
⇒pH+pOH=14 ⇒4.7+pOH=14 ⇒pOH=14−4.7 ⇒pOH=9.3 \begin{aligned} & \Rightarrow pH+pOH=14 \\\ & \Rightarrow 4.7+pOH=14 \\\ & \Rightarrow pOH=14-4.7 \\\ & \Rightarrow pOH=9.3 \\\ \end{aligned}
-Calculation of concentration of hydroxide ion present in the alcoholic drink substance as follows:-
As we know that pOH is negative logarithm of hydroxide ion i.e., pOH = −log⁡[OH−]-\log [O{{H}^{-}}]
So let us substitute the pOH value in this formula in order to calculate the concentration of hydroxide ion:-
⇒pOH=−log⁡[OH−] ⇒−log⁡[OH−]=9.3 ⇒log⁡[OH−]=−9.3 ⇒[OH−]=10−9.3 ⇒[OH−]=5×10−10 \begin{aligned} & \Rightarrow pOH=-\log [O{{H}^{-}}] \\\ & \Rightarrow -\log [O{{H}^{-}}]=9.3 \\\ & \Rightarrow \log [O{{H}^{-}}]=-9.3 \\\ & \Rightarrow [O{{H}^{-}}]={{10}^{-9.3}} \\\ & \Rightarrow [O{{H}^{-}}]=5\times {{10}^{-10}} \\\ \end{aligned}

-Therefore, the OH−O{{H}^{-}}concentration of alcoholic drink substance is (B) 5×10−105\times {{10}^{-10}}.

Note:
-Generally substances with pH less than 7 are acidic in nature whereas substances with pH greater than 7 are basic in nature. Since the pH of alcoholic drink substance is 4.7, therefore it is an acidic substance.
-Compounds or substances with pH = 7 are neutral in nature.