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3 years ago

What is the ph of a solution of 1.0 × 10-9 m naoh (include 1 digit after the decimal in your answer)?

2 answers:

6 0

Answer is: pH of NaOH solution is 5.
Reaction: NaOH → Na⁺ + OH⁻.
c(NaOH) = 1,0·10⁻⁹ mol/dm³.
pH = ?
from reaction: c(OH⁻) = c(NaOH).
c(OH⁻) = 1,0·10⁻²³ mol/dm³.
pOH = -log(c(OH⁻)) = 9.
pH + pOH = 14.
pH = 14 - pOH.
pH = 14 - 9 = 5.
pH is <span>potential of hydrogen.</span>

Eduardwww [97]3 years ago

3 0

Answer:

$pH=5$

Explanation:

Hello,

In this case, for the dissociation of the sodium hydroxide into sodium and hydroxyl ions we have:

$NaOH\rightarrow Na^++OH^-$

Which is a 100% dissociation as sodium hydroxide is a strong base, therefore, the concentration of the hydroxyl ions are 1.0x10⁻⁹. For that reason, the pOH could be computed as:

$pOH=-log([OH^-})=-log(1.0x10^{-9})=9$

Finally, from the definition of pH, we have:

$pH+pOH=14\\pH=14-pOH=14-9\\pH=5$

Best regards.

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A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

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$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$