An acidic solution has only hydrogen ions and a basic solution has only hydroxide ions statement correctly distinguishes between an acidic and a basic solution

7 0

3 years ago

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Solubility of a salt that has a general formula M3X (where M is the cation and X is the anion) is 0.0283 g/L of the solution. Th

egoroff_w [7]

$ksp = 7.003 \times {10}^{ - 16} {mol}^{4} {dm}^{ - 12}$

6 0

3 years ago

Calculate the energy required to melt 21 g of ice at 0 oC.

vladimir1956 [14]

We are given:

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

Finding the energy required:

Number of moles of Ice:

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

Energy required to melt the given amount of ice:

Energy = number of moles * molar heat of fusion

Energy = (7/6) * (6.02)

Energy = 7.02 kJ OR 7020 joules

7 0

3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$

= 0.095 mol

And, moles of NaOH present in the solution will be as follows.

No. of moles of NaOH = Molarity × Volume

= $2 \times 0.025$

= 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

$C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial: 0.095 0.05 0 0

Equlbm: (0.095 - 0.05) 0 0.05

pH = $pK_{a} + log \frac{Base}{Acid}$

= $4.2 + log \frac{0.05}{0.045}$

= 4.245

For,

$HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial: 0.2x 2(0.5 - x) 0

Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)

As,

pH = $pK_{a} + log \frac{Base}{Acid}$

4.245 = 3.75 + $log \frac{Base}{Acid}$

$log \frac{Base}{Acid}$ = 0.5

$\frac{Base}{Acid}$ = 3.162

Now,

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

= 0.036 L

= 36 ml (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

8 0

3 years ago