A solution contains 0.635 g NaOH in 236 mL solution. What is the equation for the dissociation of NaOH?
Here's what I got.
I'm guessing that the question is incomplete because you don't really need to know the mass of sodium hydroxide and the volume of the solution in order to write the chemical equation that describes the dissociation of sodium hydroxide.
Sodium hydroxide is a strong base, which implies that it dissociates completely in aqueous solution to produce sodium cations,
The balanced chemical equation that describes this dissociation looks like this
#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#
You could mention the fact that the dissociation of sodium hydroxide is water is an exothermic process, which implies that heat is being given off when sodium hydroxide dissolves in water.
This means that you can write
#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) + "heat"#
Now, you can use the data provided by the problem to determine the molarity of the solution. Molarity is a measure of the number of moles of solute present in
To convert the mass of sodium hydroxide to grams, use the molar mass of the compound
#0.635 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = "0.01588 moles NaOH"#
Next, use the volume of the solution to determine the number of moles of solute present in
#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.01588 moles NaOH"/(236color(red)(cancel(color(black)("mL solution")))) = "0.06729 moles NaOH"#
Since this represents the number of moles of solute present in
#color(darkgreen)(ul(color(black)("molarity = 0.0673 mol L"^(-1))))#
The answer is rounded to three sig figs, the number of sig figs you have for your values.
If you take into account the fact that sodium hydroxide dissociates in
#["Na"^(+)] = ["OH"^(-)] = "0.0673 mol L"^(-1)#
This will allow you to find the
#"pH" = 14 - "pOH"#
which is equivalent to
#"pH" = 14 - (- log(["OH"^(-)])#
#"pH" = 14 + log(["OH"^(-)])#
In your case, you'd have
#"pH" = 14 + log(0.0673) = 12.83#
By signing up, you agree to our Terms of Service and Privacy Policy
The equation for the dissociation of NaOH is:
NaOH (aq) → Na+ (aq) + OH- (aq)
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
- How do you calculate the osmolarity of a #6.0 * 10^-2# M NaCl solution?
- 25 mL of a #"10% m/v"# #NH_4Cl# solution is diluted to 250 mL. What is the final concentration of the #NH_4Cl#?
- What is the molality of a solution in which 15 g of #I_2# is dissolved in 500.0 g of alcohol?
- How milliliters of a 9.0 M #H_2SO_# solution are needed to make 0.35 L of a 3.5 M solution?
- When seawater evaporates, the concentration of salts increases until what happens?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7