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4 years ago

A compound has a pH of 3 what is the [H+] concentration?

Chemistry

1 answer:

Allisa [31]4 years ago

8 0

Answer: 0.001 M

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution. Mathematically, pH is expressed as

pH = -log(H+)

where H+ is hydrogen ion concentration

So, 3 = -log(H+)

(H+) = Antilog (-3)

(H+) = 0.001

Thus, [H+] concentration is 0.001 M

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Name and explain the two types of atom bonding

nordsb [41]

Ionic bonding which is the attraction between the cations(metal) and anions (non metal).

Metallic bonding is the electrostatic force of attraction between the fixed cations and the delocalized electrons.

7 0

4 years ago

Which of these is a pair of coordination isomers (aka ionization isomers)?

vagabundo [1.1K]

Answer: $[Ni(NH_3)_3Br]Cl$ and $[Ni(NH_3)_3Cl]Br$ are ionization isomers.

$[Pt(Cl)_2(SCN)_4]^{4-}$ and $[Pt(Cl)_2(NCS)_4]^{4-}$ are linkage isomers.

Explanation:

Ionization isomerism occur when a ligand that is bound to the metal center exchanges places with an anion or neutral molecule that was originally outside the coordination complex

Thus $[Ni(NH_3)_3Br]Cl$ and $[Ni(NH_3)_3Cl]Br$ are ionization isomers.

Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity of the metal to a ligand.

Thus $[Pt(Cl)_2(SCN)_4]^{4-}$ and $[Pt(Cl)_2(NCS)_4]^{4-}$ are linkage isomers.

7 0

3 years ago

Which of the following could you do to increase the strength of an electromagnet?

notsponge [240]

Change the core from wood to iron. This is because iron is a magnetic material while wood is not magnetic hence cant acquire magnetism. Other factors that would increase the strength of electromagnet would be; increasing the amount of electric current, and increasing the number of winding's.

6 0

3 years ago

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium

Slav-nsk [51]

Answer:

a) p(CO2) = 4.103 atm

p(H2) = 2.0515 atm

p(H2O) = 3.2824 atm

b) pCO2 = 3.8754 atm

pH2 = 1.8239 atm

pCO = 0.2276 atm

pH2O = 3.51 atm

c) Kp= 0.113

Explanation:

Step 1: Data given

Moles of CO2 = 0.2000 mol

Moles of H2 = 0.1000 mol

Moles of H2O = 0.1600 mol

Volume = 2.000 L

Temperature = 500 k

Step 2: The balanced equation

CO2 (g) + H2 (g) → CO (g) + H2O (g)

Step 3: Calculate the initial partial pressures of CO2, H2, and H2O.

pV = nRT

P = (nRT)/V

p(CO2) = (0.2000 mol * 0.08206 * 500)/2.000L

⇒ p(CO2) = 4.103 atm

p(H2) = (0.1000 mol * 0.08206 * 500)/2.000L

⇒ p(H2) = 2.0515 atm

p(H2O) = (0.1600 mol * 0.08206 * 500)/2.000L

⇒ p(H2O) = 3.2824 atm

b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.

Step 1: Calculate the change in pH2O

The change in pH2O = 3.51 - 3.2824 = 0.2276

Step 2: the initial pressures

pCO2 = 4.103 atm

pH2 = 2.0515 atm

pCO = 0 atm

pH2O = 3.2824

Step 3: The partial pressure at the equilibrium

Since there reacts 0.2276 atm for H2O, and the mol ratio is 1:1:1:1

For each gas, there will react 0.2276 atm

pCO2 = 4.103 - 0.2276 = 3.8754 atm

pH2 = 2.0515 - 0.2276 = 1.8239 atm

pCO = 0 + 0.2276 = 0.2276 atm

pH2O = 3.51 atm

c. Calculate Kp for this reaction

Kp = (pCO * pH2O)/ (pCO2 * pH2)

Kp = (0.2276 * 3.51) /( 3.8754 *1.8239)

Kp = 0.113

5 0

3 years ago

An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that $pH = -log[H^{+}]$ , and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

$pH = pKa - log \frac{[HA]}{[A^{-}]}$

Where [HA] is the concentration of the acid, and $[A^{-}]$ is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of $OH^-$ will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of $OH^-$ reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of $OH^-$ , which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

$6.72 = pKa - log \frac{0.01}{0.0025}$

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0

3 years ago