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3 years ago

A solution with [ OH] of 5 x 10-3 has a pH of

Chemistry

1 answer:

CaHeK987 [17]3 years ago

8 0

Answer:

11·699

Explanation:

Given the concentration of hydroxide ion in the solution is 5 × $10^{-3}$ M

Assuming the temperature at which it is asked to find the pH of the solution be 298 K

<h3>At 298 K the dissociation constant of water is $10^{-14}$ </h3><h3>∴ pH + pOH = 14 at 298 K</h3><h3>pOH of the solution = -log( concentration of hydroxide ion )</h3>

∴ pOH of the given solution = - log(5 × $10^{-3}$ = -0·699 + 3 = 2·301

pH of the given solution = 14 - 2·301 = 11·699

∴ pH of the solution = 11·699

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We make a basic solution by mixing 50. mL of 0.10 M NaOH and 50. mL of 0.10 M Ca(OH)2. It requires 250 mL of an HCl solution to

andreyandreev [35.5K]

<u>Answer:</u> The correct answer is Option 5.

<u>Explanation:</u>

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.10M\\V_1=50mL\\n_2=2\\M_2=0.1\\V_2=50mL$

Putting all the values in above equation, we get:

$M=\frac{(1\times 0.1\times 50)+(2\times 0.1\times 50)}{50+50}\\\\M=0.15M$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base.

We are given:

$n_1=1\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.15M\\V_2=100mL$

Putting values in above equation, we get:

$1\times M_1\times 250=1\times 0.15\times 100\\\\M_1=0.06M$

Hence, the correct answer is Option 5.

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Answer:

Explanation:

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