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amid [387]
2 years ago
11

What is the pH of a solution of 1 x 10-5 M NaOH? Is the solution acidic or basic?

Chemistry
1 answer:
Anna007 [38]2 years ago
3 0

Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.

Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9

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2 years ago
he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o
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<u>Answer:</u> The number of moles of HI in the solution is 1.24\times 10{-3} moles.

<u>Explanation:</u>

We are given:

K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L

To calculate the concentration of a substance, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     ......(1)

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[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L

  • Concentration of ammonium iodide:

[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L

For the given chemical reaction:

NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)

The expression of K_c for above equation follows:

K_c=\frac{[HI][NH_3]}{[NH_4I]}

Putting values in above equation, we get:

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Calculating the moles of hydrogen iodide by using equation 1, we get:

2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}

Hence, the number of moles of HI in the solution is 1.24\times 10{-3} moles.

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