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Question: Calculate pH and pOH of \( 1.0 \times {10^{ - 7}} \) HCl solution? (Hint: use quadratic equation)....

Calculate pH and pOH of 1.0×10−71.0 \times {10^{ - 7}} HCl solution? (Hint: use quadratic equation).

Explanation

Solution

The pH of any solution can be calculated by the formula: pH=−log⁡[H3O]+pH = - \log {[{H_3}O]^ + }
If the concentration of the solution is below or equal to 10−7M{10^{ - 7}}M we’ll have to consider the hydronium ion concentration of the water also. Hence the concentration of Hydronium ions will be the sum of hydronium ions from HCl and hydronium ions from water. The value of pOH can be found out by the pH itself as, pOH=14−pHpOH = 14 - pH .

Complete answer:
We are given that the concentration of acid is 1.0×10−71.0 \times {10^{ - 7}} , this concentration is very low. Hence, we’ll have to dissociate water also instead of directly taking the [H+][{H^ + }] concentration as 10−7M{10^{ - 7}}M . The dissociation of water can be given as:
H2O⇌H++OH−{H_2}O \rightleftharpoons {H^ + } + O{H^ - }
The equilibrium constant Kw=[H+][OH−]=10−14{K_w} = [{H^ + }][O{H^ - }] = {10^{ - 14}}
The addition of HCl might have slightly changed the equilibrium increasing the [H+][{H^ + }] concentration. The system will try to shift the equilibrium backwards and given a new equilibrium. Therefore, the initial concentration of [H+][{H^ + }] from water to be 10−7M{10^{ - 7}}M + 10−7M{10^{ - 7}}M [H+][{H^ + }] from HCl which makes a total of [H+]=2×10−7M[{H^ + }] = 2 \times {10^{ - 7}}M
The equilibrium now can be given as:
H2O ⇌ H+ + OH−{H_2}O{\text{ }} \rightleftharpoons {\text{ }}{H^ + }{\text{ }} + {\text{ }}O{H^ - }

T=02×10−72 \times {10^{ - 7}}10−7{10^{ - 7}}
T=equilibrium2×10−7−x2 \times {10^{ - 7}} - x10−7−x{10^{ - 7}} - x

Therefore, the equilibrium constant now can be written as, Kw=[2×10−7−x][10−7−x]=10−14{K_w} = [2 \times {10^{ - 7}} - x][{10^{ - 7}} - x] = {10^{ - 14}}
On opening the brackets, we get, x2−(3×10−7)x+10−14=0{x^2} - (3 \times {10^{ - 7}})x + {10^{ - 14}} = 0
On applying the quadratic formula and ignoring the roots we get the value of x as: x=0.385×10−7mol/Lx = 0.385 \times {10^{ - 7}}mol/L
The concentration of [H+][{H^ + }] = 2×10−7−x2 \times {10^{ - 7}} - x =2×10−7−0.385×10−7=1.615×10−7mol/L= 2 \times {10^{ - 7}} - 0.385 \times {10^{ - 7}} = 1.615 \times {10^{ - 7}}mol/L
Therefore, the value of pH will be =−log⁡[H+]=−log⁡(1.615×10−7)=6.76= - \log [{H^ + }] = - \log (1.615 \times {10^{ - 7}}) = 6.76
Hence pH = 6.79
Since we know the pH, we can find the pOH as: pOH=14−6.76=7.208pOH = 14 - 6.76 = 7.208
Our required answer for pH and pOH is 6.76 and 7.208 respectively.

Note:
If we do not consider the ionisation of water and ignore the [H+][{H^ + }] from water we’ll get the answer as: pH=−log⁡(10−7)=7pH = - \log ({10^{ - 7}}) = 7 , which gives us neutral solution. Acidic solution will always have pH<7pH < 7 . Hence the answer obtained would be incorrect.