What is the pH of $9.55 \times 10^{-7} M HCl \left( H ^{+}\right)$ solution?
A. 9.55
B. 0.166
C. $1.05 \times 10^{-8}$
D. 6.02
E. none of these available choices
Answer (2)
The pH of the 9.55 × 1 0 − 7 M HCl solution is approximately 6.02, considering the complete dissociation of HCl and the autoionization of water. The total hydrogen ion concentration is 1.2565311 × 1 0 − 6 M . Therefore, the correct answer is 6.02.
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Determine the concentration of H + ions from H Cl : [ H + ] H Cl = 9.55 × 1 0 − 7 M .
Consider the autoionization of water and set up the equation: ( 9.55 × 1 0 − 7 + x ) ( x ) = 1.0 × 1 0 − 14 .
Solve the quadratic equation to find the concentration of H + from water: x = 3.015311 × 1 0 − 7 M .
Calculate the total concentration of H + and find the pH: p H = − lo g 10 ( 1.2565311 × 1 0 − 6 ) ≈ 6.02 . The final answer is 6.02 .
Explanation
Initial Analysis We are given the concentration of H Cl which is $9.55
\times 10^{-7} M . S in ce HCl i s a s t ro n g a c i d , i t co m pl e t e l y d i ssoc ia t es in w a t er . T h ere f ore , t h eco n ce n t r a t i o n o f H^+ i o n s f ro m HCl i se q u a lt o t h eco n ce n t r a t i o n o f t h e HCl so l u t i o n , w hi c hi s [H^+]_{HCl} = 9.55
\times 10^{-7} M$.
Considering Water Autoionization However, we must also consider the autoionization of water, which contributes to the H + concentration. The autoionization of water is represented by the equilibrium: $H_2O
\rightleftharpoons H^+ + OH^- . T h e i o n p ro d u c t o f w a t er , K_w , i s g i v e nb y K_w = [H^+][OH^-] = 1.0
\times 10^{-14} a t 25° C . S in ce t h eco n ce n t r a t i o n o f HCl$ is close to 1 0 − 7 M , we cannot ignore the contribution of water to the H + concentration.
Calculating H+ from Water Let x be the concentration of H + from water. Then, the total concentration of H + is $[H^+]_{total} = 9.55
\times 10^{-7} + x . T h eco n ce n t r a t i o n o f OH^- i s a l so x$. Therefore, we have:
( 9.55 × 1 0 − 7 + x ) ( x ) = 1.0 × 1 0 − 14
Expanding this equation, we get:
x 2 + 9.55 × 1 0 − 7 x − 1.0 × 1 0 − 14 = 0
This is a quadratic equation in the form of a x 2 + b x + c = 0 , where a = 1 , $b = 9.55
\times 10^{-7} , an d c = -1.0
\times 10^{-14} . W ec an so l v e f or x$ using the quadratic formula:
x = 2 a − b ± b 2 − 4 a c
x = 2 ( 1 ) − 9.55 × 1 0 − 7 ± ( 9.55 × 1 0 − 7 ) 2 − 4 ( 1 ) ( − 1.0 × 1 0 − 14 )
x = 2 − 9.55 × 1 0 − 7 ± 9.12025 × 1 0 − 13 + 4.0 × 1 0 − 14
x = 2 − 9.55 × 1 0 − 7 ± 13.12025 × 1 0 − 13
x = 2 − 9.55 × 1 0 − 7 ± 3.622188 × 1 0 − 7
Since x must be positive, we take the positive root:
x = 2 − 9.55 × 1 0 − 7 + 3.622188 × 1 0 − 7 = 2 − 5.927812 × 1 0 − 7 = 3.015311 × 1 0 − 7
Calculating Total H+ and pH Therefore, $x = 3.015311
\times 10^{-7} M . T h e t o t a l co n ce n t r a t i o n o f H^+$ is:
[ H + ] t o t a l = 9.55 × 1 0 − 7 + 3.015311 × 1 0 − 7 = 12.565311 × 1 0 − 7 = 1.2565311 × 1 0 − 6 M
Now, we can calculate the pH using the formula:
p H = − lo g 10 [ H + ]
p H = − lo g 10 ( 1.2565311 × 1 0 − 6 )
p H ≈ 6.015311
Final Answer The pH of the solution is approximately 6.02. Therefore, the correct answer is 6.02.
Examples
Understanding pH is crucial in many real-world applications. For instance, in agriculture, knowing the pH of the soil helps farmers determine the best crops to grow and the necessary soil treatments. In medicine, maintaining the correct pH balance in the body is essential for various biological processes. Additionally, in environmental science, monitoring the pH of water sources helps assess pollution levels and ensure water quality. The calculation of pH, as demonstrated in this problem, is a fundamental skill in these fields.
