Question

Calculate the pH of 10-6 M HCl solution. When 10-6 M HC and 10-5 M NaOH solutions are mixed
in equal amounts, what will be the pH of the mixed solution?

Answer 1

pH of 10⁻⁶ M HCl is 6

pH of the mixed solution is 8.954

Explanation:

We know that

$pH=-\log[H^+]$

Given concentration of $H^+$ ions in HCl solution

$[H^+]=10^{-6}$ mole/litre

Therefore, the pH of HCl solution

$pH=-\log 10^{-6}$

or, $pH=6$

When 10⁻⁶ M HCl is mixed with 10⁻⁵ M NaOH then the reaction is

HCl + NaOH ⇒ NaCl + H₂O

1 mole of HCl reacts with 1 mole of NaOH

Therefore, 10⁻⁶ mole HCl will react with 10⁻⁶ mole NaOH

Remaining moles of NaOH

$=10^{-5}-10^{-6}$

$=10^{-6}(10-1)$

$=9\times 10^{-6}$

Therefore concentration of OH⁻ ions

$[OH^-]=9\times 10^{-6}$ mole/litre

We know that

$[H^+][OH^-]=10^{-14}$

$\implies [H^+]=\frac{10^{-14}}{[OH^-]}$

$\implies  [H^+] = \frac{10^{-14}}{9\times 10^{-6}}$

$\implies [H^+] =1.11\times 10^{-9}$

Therefore, the pH of the solution is

$pH=-\log(1.11\times 10^{-9}$

$\implies pH=9-0.046$

$\implies pH=8.954$

Hope this answer is helpful.

Know More:

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Q: Calculate the pH of 0.001M NaOH.Assuming complete dissociation of base.

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Answer 2

Answer:

pH of 10⁻⁶ M HCl is 6

pH of the mixed solution is 8.954

Explanation:

We know that

pH=-\log[H^+]pH=−log[H

+

]

Given concentration of H^+H

+

ions in HCl solution

[H^+]=10^{-6}[H

+

]=10

−6

mole/litre

Therefore, the pH of HCl solution

pH=-\log 10^{-6}pH=−log10

−6

or, pH=6pH=6

When 10⁻⁶ M HCl is mixed with 10⁻⁵ M NaOH then the reaction is

HCl + NaOH ⇒ NaCl + H₂O

1 mole of HCl reacts with 1 mole of NaOH

Therefore, 10⁻⁶ mole HCl will react with 10⁻⁶ mole NaOH

Remaining moles of NaOH

=10^{-5}-10^{-6}=10

−5

−10

−6

=10^{-6}(10-1)=10

−6

(10−1)

=9\times 10^{-6}=9×10

−6

Therefore concentration of OH⁻ ions

[OH^-]=9\times 10^{-6}[OH

−

]=9×10

−6

mole/litre

We know that

[H^+][OH^-]=10^{-14}[H

+

][OH

−

]=10

−14

\implies [H^+]=\frac{10^{-14}}{[OH^-]}⟹[H

+

]=

[OH

−

]

10

−14

\implies [H^+] = \frac{10^{-14}}{9\times 10^{-6}}⟹ [H

+

]=

9×10

−6

10

−14

\implies [H^+] =1.11\times 10^{-9}⟹[H

+

]=1.11×10

−9

Therefore, the pH of the solution is

pH=-\log(1.11\times 10^{-9}pH=−log(1.11×10

−9

\implies pH=9-0.046⟹pH=9−0.046

\implies pH=8.954⟹pH=8.954

Hope this answer is helpful.

Know More:

Q: Calculate the pH of 1.0 ×10–4 molar solution of HNO3.

Click Here: https://brainly.in/question/11844823

Q: Calculate the pH of 0.001M NaOH.Assuming complete dissociation of base.

Click Here: https://brainly.in/question/14887815