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3 years ago

Which is one way to test whether an unknown solution is acidic or basic

Chemistry

2 answers:

lions [1.4K]3 years ago

8 0

Testing the ph. above 7 is acid and below is base. hope that helps

yarga [219]3 years ago

7 0

if the ph is above 7 then its acidic

if its below 7 than its going to be basic.

~serenity bella

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You want to make 500 ml of a 1 N solution of sulfuric acid (H2SO4, MW: 98.1). How many grams of sulfuric acid do you need?

umka21 [38]

Answer:

24.525 g of sulfuric acid.

Explanation:

Hello,

Normality (units of eq/L) is defined as:

$N=\frac{eq_{solute}}{V_{solution}}$

Since the sulfuric acid is the solute, and we already have the volume of the solution (500 mL) but we need it in liters (0.5 L, just divide into 1000), the equivalent grams of solute are given by:

$eq_{solute}=N*V_{solution}=1\frac{eq}{L}*0.5L=0.5 eq$

Now, since the sulfuric acid is diprotic (2 hydrogen atoms in its formula) 1 mole of sulfuric acid has 2 equivalent grams of sulfuric acid, so the mole-mass relationship is developed to find its required mass as follows:

$m_{H_2SO_4}=0.5eqH_2SO_4(\frac{1molH_2SO_4}{2 eqH_2SO_4}) (\frac{98.1 g H_2SO_4}{1 mol H_2SO_4} )\\m_{H_2SO_4}=24.525 g H_2SO_4$

Best regards.

4 0

3 years ago

Hey how's ur friday going yall im bacc in school now so yay!

Naily [24]

My Friday is going good and I might get to go over to my BFF house!!!

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2 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago

A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH

xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

6 0

3 years ago

P= f/a solve it for f

Bess [88]

Answer:

Force = Pressure × Area

Explanation:

Easy, just invert the equation. Transpose the force variable over to the left of the equals sign, and transpose the pressure variable back to the right side.

7 0

2 years ago