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Crank
3 years ago
10

What is the pH of 1.0M HI solution?

Chemistry
2 answers:
ozzi3 years ago
7 0

Answer:

0

Explanation:

Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.

ELEN [110]3 years ago
4 0

Answer:

<h2>pH = <u>0</u></h2>

Explanation:

{hope it helps}}

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The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the con
Kobotan [32]

Answer:

The correct answer is 190.5 mL of 1.00 M KH₂PO₄

Explanation:

A phosphate buffer is composed by phosphate acid (KH₂PO₄) and its conjugated base (K₂HPO₄). To obtain the relation between the concentrations of base and acid to add, we use Henderson-Hasselbach equation:

pH= pKa + log \frac{base}{acid}

We have: pH= 6.97 and pKa= 7.21. So, we replace the values in the equation:

6.97= 7.21 + log \frac{base}{acid}

6.97-7.21= log \frac{base}{acid}

-0.24= log \frac{base}{acid}

10^{-0.24}= \frac{base}{acid}

0.575 = \frac{base}{acid}

\frac{0.575}{1}= \frac{base}{acid}

It means that you have to mix a volume 0.575 times of conjugated base and 1 volume of acid. If we assume a total buffer concentration of 1 M, we have:

base + acid = 1

base= 1 - acid

We replace in the previous equation:

0.575= \frac{1-acid}{acid}

0.575 acid= 1 - acid

0.575 acid + 1 acid= 1

1.575 acid = 1

acid= 1/1,575

acid= 0.635

base= 1 - acid = 1 - 0.635 = 0.365

For a total volume of 300 ml, the volumes of both acid and base will be:

300 ml x 0.635 M = 190.5 ml of acid (KH₂PO₄)

300 ml x 0.365 M= 109.5 ml of base (K₂HPO₄)

We can corroborate our calculations as follows:

190.5 ml + 109.5 ml = 300 ml

109.5 ml / 190.5 ml = 0.575

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Answer:

I can, i love chemistry

Explanation:

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3 years ago
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19) What is the molarity of a KOH solution if 200 ml of the solution contains 0.6 moles KOH?
zloy xaker [14]

Answer: 3M

Explanation: Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute KOH = 0.6 moles

V_s = volume of solution in ml= 200 ml

Now put all the given values in the formula of molarity, we get

Molarity=\frac{0.6moles\times 1000}{200ml}=3mole/L

Therefore, the molarity of solution will be 3M.

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