Calculate [OH^-] and pH for each of the following strong base solutions: (a) 0.182 M KOH, (b) 3.

step 1 Chapter 16, problem 46, from chemistry, the essential science. So we're going to be calculating

Calculate [OH^-] and pH for each of the following strong...

Key Concepts

Strong Bases

Strong bases are compounds that fully dissociate into their ions in aqueous solution, meaning that the concentration of hydroxide ions produced is directly related to the initial concentration of the base. This complete dissociation simplifies the calculation of [OH?] as no equilibrium considerations are needed.

Molarity

Molarity is a concentration unit defined as moles of solute per liter of solution. It is fundamental for determining the number of moles of a substance present and is used to calculate the resulting [OH?] in solutions of strong bases.

Stoichiometry in Dissociation

Stoichiometry in dissociation refers to the mole ratios in which a compound ionizes in solution. For strong bases, it is crucial to consider how many hydroxide ions are produced per formula unit, as some bases yield more than one OH?, affecting the overall ion concentration in the solution.

Dilution Principles

Dilution involves adding solvent to reduce the concentration of a solution. The relationship M?V? = M?V? is used to calculate the new concentration when a solution is diluted, which directly influences the [OH?] and subsequent pH value.

pH and pOH Relationship

pH and pOH are related measures of acidity and basicity in a solution, respectively. The formula pH + pOH = 14 (at 25°C) allows for the calculation of one value when the other is known, enabling the determination of pH from the hydroxide concentration calculated.

Transcript

00:02 Chapter 16, problem 46, from chemistry, the essential science.

00:08 So we're going to be calculating the ph from oh concentrations.

00:16 So for a, they gave us a oh concentration of 0 .182 molarity of k -o -h.

00:29 So what we're first going to do is going to calculate our p -o -h from this concentration, which is the negative log of our o -h concentration.

00:40 And when i plug this into my calculator, i get the value of 0 .74.

00:53 And so this is my p -o -h, the 7 .4.

01:00 But we're running our ph for this equation.

01:03 So to get my ph, all we're going to do is 14 is equal to our ph plus our poh into 14 minus 0 .74 will give me 13 .26.

01:18 That's our answer for a.

01:22 So for question, part b, they give us a solution, mean they give us three.

01:34 0 .165 grams of k -o -h and they dissolve it in 5 -mil -liter's solution.

01:43 So what we need to do is we need to find the molarity of the solution and molarity is equal to your moles over your leaders.

01:52 And we know what our leaders is and we can do that by converting 500, by converting 500 milliliters and two liters because there's a thousand milliliters and one liter and that gives a 0 .5 liters.

02:20 Also we need to calculate the moles of k -o -h and to do that we need to find the molecular weight of k -o -h and to do that we just look at our periodic table and find the molecular weight of one potassium plus one oxygen plus one hydrogen.

02:43 And so the total molecular weight for k -o -h is 56 .11 grams per mole.

02:58 So now since we have the molecular weight, we can find the moles of k -o -h -given.

03:07 Due to the amount of grams.

03:08 And to do that, we just do grams divided by the molecular weight, which is 3 .165 grams divided by 56 .11 grams per mole.

03:23 And we get 0 .0564 moles of k -o -h.

03:41 Now, since we have our moles and we have our volume, we just do 0 .0564 divided by 0 .5 .5.

03:52 And we get our concentration of our oh.

04:06 So now since we have our concentration of our oh, we can find our p -o -h and using that same equation that we used above.

04:27 And when i plug it into my calculator, i get a value of.

04:38 0 .96 is what my p .o .h equals to.

04:44 So to find our p .h.

04:47 It is just 14 minus 0 .96.

04:52 And so my p .h is equal to 13 .04.

04:59 So we have now solved a and b.

05:02 So question c, they give us 10 milliliters of 0 .015 molar of c .a.

05:13 Oh, h2, diluted to 500 milliliters.

05:20 And so here we had to figure out how much oh concentration we actually have in this solution.

05:26 So first we're going to do a m1v1 solution to figure out what our dilution of our, concentration actually is.

05:39 And so m2 is equal to m1 v1 over b2, because this is what we're trying to solve for.

05:48 And so our m1 is 0 .0105 molar.

05:58 Our initial volume is 10 milliliters.

06:02 And then we dilute it to 500 milliliters.

06:10 All right.

06:10 So when i plug that into my calculator, i get a concentration of 2 .10 times 10 to the negative 4 molarity.

06:22 All right.

06:23 So now since i know the malaria of the caoh2, now i can figure out what my oh concentration is, but i have to be careful because i have two o .hs in this.

06:43 So we need to figure out how many ohs we have per hydrogen ion when we're trying to convert back to our ph.

06:56 So to do that, we just need to break up our ion solution...

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