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3 years ago

What are the products produced when an acid and base react together

Chemistry

2 answers:

SashulF [63]3 years ago

4 0

Both products will start to cancel the acidity and how strong the base is if they are mixed. If the acid is stronger than the base then it will be an acidic product and visa versa if the base is stronger than the acid.

sattari [20]3 years ago

4 0

But remember that water is a mixed of acid and base together.
Always look at the PH scale if you’re not sure ;)

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Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy for the following redox

stealth61 [152]

Answer:

-973 KJ

Explanation:

The balanced reaction equation is;

N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)

Reduction potential of hydrazine = -1.16 V

Reduction potential of chlorine = 1.36 V

From;

E°cell= E°cathode - E°anode

E°cell= 1.36 - (-1.16)

E°cell= 2.52 V

∆G°=- nFE°cell

n= number of moles of electrons = 4

F= Faraday's constant = 96500 C

E°cell = 2.52 V

∆G°=- (4 × 96500 × 2.52)

∆G°= -972720 J

∆G°= -972.72 KJ

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3 years ago

The highest energy level of sodium contains ________ electron(s).

Effectus [21]

Answer:

Explanation:

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3 years ago

Electronegativities of the elements Be, Mg, Ca, and Sr follow a specific trend within their group. Based on this trend, the atom

Travka [436]

Answer:

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2 years ago

Read 2 more answers

Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)

klio [65]

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

Best regards.

8 0

3 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0

3 years ago