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3 years ago

How is the pH of an acidic substance raised?

Chemistry

2 answers:

Dvinal [7]3 years ago

8 0

Answer:

Concentrate it.

Explanation:

I got it right on the test.

ch4aika [34]3 years ago

4 0

Explanation:

An acid is a substance that produces excess hydroxonium ions in solution.

An acid based on the pH scale is a substance that has a low pH. Acid lies within a range of 1-7 on the pH scale.

A pH of 7 is for neutral compounds like water.
A pH greater that 7 is for basic compounds.
In order to raise the pH, we are driving at a substance becoming more neutral or basic.

This can be achieved by adding more base to the solution of the substance. When we add more base, hydroxyl ions will neutralize the excess hydroxonium ions and drag the pH towards that of neutrality.

Addition of more base can eventually make the substance basic.

learn more:

Phosphoric acid brainly.com/question/11062486

#learnwithBrainly

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Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

The empirical formula of styrene is ch; its molar mass is 104.1 g/mol. what is the molecular formula of styrene? select one:

Nuetrik [128]

The empirical formula is a formula of a compound showing the proportion of each element involved in the compounds but it does not represent the total number of atoms in the compound. It is the lowest number of ratio between the elements in the compound. In order, to determine the actual number of the atoms or the molecular formula of the compounds, we make use of the molar mass of the compound.

To determine the molecular formula, we multiply a value to the empirical formula. Then, calculate the molar mass and see whether it is equal to the one given (104.1 g/ mol). From the choices, the only valid options are b, d and e.
 molar mass
1 CH 13.02
8 C8H8 104.16
6 C6H6 78.12

Therefore the correct answer is option B.

6 0

3 years ago

Read 2 more answers

Calculate # of atoms in 46g of Na

makvit [3.9K]

21Explanation:BEACUSE I CAN

5 0

2 years ago

Which of these has the most covalent character? NaF Caf2 Bef2 CsF SrF2

Ksenya-84 [330]

Thus BeF2 is of most covalent character.

Anyways, covalent/ionic character is a bit tricky to figure out; we measure the difference in electronegativity of two elements bonding together and we use the following rule of thumb: if the charge is 0 (or a little more), the bond is non-polar covalent; if the charge is > 0 but < 2.0 (some references say 1.7), the bond is polar covalent; if the charge is > 2.0 then the bond is ionic. Covalent character refers to smaller electronegativity difference while ionic character refers to greater electronegativity difference.

Now, notice all of our bonds are with F, fluorine, which has the highest electronegativity of 3.98. This means that to determine character we need to consider the electronegativities of the other elements -- whichever has the greatest electronegativity has the least difference and most covalent character.
Na, sodium, has electronegativity of 0.93, so our difference is ~3 -- meaning our bond is ionic. Ca, calcium, has 1.00, leaving our difference to again be ~3 and therefore the bond is ionic. Be, beryllium, has 1.57 yielding a difference of ~2.5, meaning we're still dealing with ionic bond. Cs, cesium, has 0.79, meaning our difference is again ~3 and therefore again our compound is of ionic bond. Lastly, we have Sr, strontium, with an electronegativity of 0.95 and therefore again a difference of roughly 3 and an ionic bond.

3 0

3 years ago

Consider the following reaction:

adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

2H₂O₂ ⟶ 2H₂O + O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂

(a) Initial moles of H₂O₂

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }$

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

$\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }$

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

$\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}$

8 0

3 years ago