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Gemiola [76]
3 years ago
12

Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd

rochloric acid. The pka values for phosphoric acid are 2.1, 7.2 and 12.4.
Chemistry
1 answer:
GenaCL600 [577]3 years ago
7 0

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4  + H+  ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of  H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4  + H+  ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

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Explanation:

local winds are considered breezes while global winds produce mostly storms

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What is the volume of a sample of ammonia gas (NH3) if it contains 0.23 moles?​
LekaFEV [45]

The volume of a sample of ammonia gas : 5.152 L

<h3>Further explanation</h3>

Given

0.23 moles of ammonia

Required

The volume of a sample

Solution

Assumed on STP

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

So for 0.23 moles :

= 0.23 x 22.4 L

= 5.152 L

8 0
2 years ago
A very flammable substance would not be used in children’s clothing? true or false
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True because certain fabric will catch on fire 
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3 years ago
A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

7 0
3 years ago
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0
3 years ago
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