What is the [H+] if pOH =9.50
will it be acid or basic?
Taking into account the definition of pH and pOH, the [H⁺] is 3.16×10⁻⁵ M and the solution will be acid.
First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.
The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:
pH= - log [H⁺]= - log [H₃O⁺]
Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
The following relationship can be established between pH and pOH:
pH + pOH= 14
In this case, being pOH= 9.50, pH is calculated as:
pH + 9.40= 14
pH= 14 - 9.50
pH= 4.50
Replacing in the definition of pH the concentration of H⁺ ions is obtained:
- log [H⁺]= 4.50
Solving :
[H⁺]= 10⁻⁴ ⁵
[H⁺]= 3.16×10⁻⁵ M
The numerical scale that measures the pH of the substances includes the numbers from 0 to 14, being acidic solutions with a pH lower than 7, and basic those with a pH greater than 7. The pH = 7 indicates the neutrality of the solution.
In this case, the pH has a value of 4.50. So, the solution is acidic.
In summary, the [H⁺] is 3.16×10⁻⁵ M and the solution will be acid.
Learn more:
brainly.com/question/16032912?referrer=searchResults brainly.com/question/13557815?referrer=searchResultsWhat identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!
Oxidation unit test
Please help ASAP!!!
Which statement correctly describes the oxidation number of the manganese atom (Mn) in Mnl2 and MnO2?
O Manganese has an oxidation number of +4 in Mnl2 and +2 in MnO2.
o Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
o Manganese has an oxidation number of +4 in both Mnl2 and MnO2.
Manganese has an oxidation number of +2 in both Mnl2 and MnO2.
In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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https://brainly.com/question/15167411 https://brainly.com/question/6710925Determine the molarity for each of the following Solution: 98.0 of phosphoric acids H3PO4 in 1.00L of Solution.
The molarity of the solution is 0.01.
Brainliest?
In a constant-pressure calorimeter, 55.0 mL of 0.340 M Ba(OH), was added to 55.0 mL of 0.680 M HCI. The reaction caused the temperature of the solution to rise from 22.21 °C to 26.84 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g • °C.) respectively), what is A4 for this reaction (per mole H,O produced)? Assume that the total volume is the sum of the individual volumes.
Answer:
Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O
The reactants are present in equimolar amounts, so there is no excess or limiting reactants.
(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O
(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J
(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O
Explanation:
Find five household items that contain ionic compounds. List the following information for each compound: a.name of the product
b. the chemical name of the ionic compound
d.the chemical formula of the ionic compound c. identify the ions present in each compound
Answer:
D . GOD BLESS .
Explanation:
a. Calculate the change in enthalpy when 20.0 grams of aluminum metal is heated from 298 K to 573 K at constant pressure of 1 atm.
b. Calculate the change in enthalpy when 20.0 grams of metallic lead is taken through the same process. In both cases assume the heat capacity values predicted by equipartition are valid through the temperature range stated.