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3 years ago

What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

Chemistry

1 answer:

AlekseyPX3 years ago

5 0

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

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What is a compounds

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3 years ago

hat is the pressure of CO(g) in equilibrium with the CO2(g) and O2(g) in the atmosphere at 25 ????C? The partial pressure of O2(

Lubov Fominskaja [6]

Answer:

The partial pressure of CO is 5.54x10⁻⁴⁹atm. You shouldn't worry because it is very low pressure

Explanation:

First, the balanced reaction is:

CO + 1/2O₂ → CO₂

The energies of formation are:

ΔG(CO)=-137.168kJ/mol

ΔG(O₂)=0

ΔG(CO₂)=-394.359kJ/mol

The energy of the reaction is:

$delta-G_{reaction} =delta-G_{CO_{2} } -(delta-G_{CO} +1/2delta-G_{O_{2} } )\\delta-G_{reaction}=-394.359-(-137.168+0)=-257.191kJ/mol$

The expression for calculate the partial pressure of CO is:

$p_{CO} =\frac{p_{CO2} }{p_{O_{2} }^{1/2}*exp^{-\frac{delta-G}{RT} } } \\p_{CO}=\frac{3x10^{-4} }{0.2^{1/2}*exp(-\frac{-257.191*1000}{8.314*298} ) } \\p_{CO}=5.54x10^{-49} atm$

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3 years ago

How many moles of H2 can be made from complete reaction of 3.0 moles of Al?

tresset_1 [31]

Answer: 4.5 moles of $H_{2}$ can be made from complete reaction of 3.0 moles of Al.

Explanation:

The given reaction equation is as follows.

$2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}$

This shows that 2 moles of Al reacts with 6 moles of HCl. So, the amount of HCl required to react with 1 mole Al is three times the amount of HCl.

Therefore, 3 moles of Al will react with 9 moles of HCl to give 3 moles of $AlCl_{3}$ and $\frac{9}{2}$ moles of $H_{2}$ .

The reaction equation now will be as follows.

$3Al + 9HCl \rightarrow 3AlCl_{3} + \frac{9}{2}H_{2}$

The moles $\frac{9}{2}$ can also be written as 4.5 moles.

Thus, we can conclude that 4.5 moles of $H_{2}$ can be made from complete reaction of 3.0 moles of Al.

6 0

3 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

Answer:

For (a): The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

For (b): The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

For (c): The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

For (d): The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

For (e): The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

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3 years ago

Graciella divided her grapes equally among 6 friends. If each friend received 16 grapes how many grapes did Graciella have?

DIA [1.3K]

Answer: She had 96 grapes.

Explanation:

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3 years ago

Read 2 more answers