Chemistry problem - in-depth help needed?

Question

Calculated the pH of the following solutions.
a) 5.00 g of HBr in 100 mL of aqueous solution
b) 1.50 g of NaOH in 50 mL of aqueous solution

How do I calculate the pH? What are the steps I need to take?

Answer 1

a) First, it is important to know that HBr is a strong acid. That is, it completely dissolves into H+ and Br- ions. Now, we can find the molarity concentration of the H+ ions. HBr has a molar mass of 80.91 g/mol, so 5.00 grams creates .0618 mol of HBr. Molarity is calculated by dividing mols of solute by liters of solution. In that case, molarity is .0618 mols / .100 L = .618 M. Finally, pH = -log[molarity of H+], which is .209.

b) To calculate pH of a base, first calculate the pOH. NaOH is a strong base, so it completely dissolves. We know that the molar mass of NaOH is 40.0 g/mol, so 1.50 g gives us .0375 moles of OH- ions. The molarity is .0375 mols / .050 L = .75 M. The pOH = -log[molarity of OH-] = .125. The sum of the pH and pOH of any substance always equals 14, so our pH is 14 - .125 = 13.875.

Answer 2

It's been a while but...isn't the pH equation: pH= -log [ H+], where [ ] is the concentration of H+ ions in Molar units?

I'd go with

a) 5.00g/100ml = 0.05 g/mL and then change it into g/L for M which would be 50 g/L or M
so pH of Hbr (which is an acid) = -log (50) = 1.70

For b) however, NaOH is a base, so you'd have to follow the same step as above, but since that is a base, you'd be finding the pOH, so get the answer and subract from 14 because
pH + pOH =14

so for b) 1.50g/50ml = 0.30 g/mL which then is 30 g/ L or M
so pOH = -log (30) = 1.47
so taking the formula pH + pOH = 14 you could then get the following:

pH = 14 - pOH
pH = 14 - 1.47 = 12.53

I hope I didn't confuse you more! It's been a couple years since my chem classes

Answer 3

pH value is the negative logarithm of the concentration of H+ ions in the solution. The concentration is measured in moles/Litre. Calculate the molar weight of HBr in grams. So that will give you how many moles you have in 5 g. This is in 100mL of water.. easy to convert to moles/Liter. Then take the logarithm to base 10 of the concentration and make the number negative.
Similar steps for the NaOH solution.

Most previous answerers have done the problem for you. I hope this is not a homework problem.

Answer 4

I cannot do both questions because there are no Ka or Kb values given. The compounds given above dissociate only to a certain degree in an aqueous environment, and that degree is found by using a Ka or Kb value, depending on the compound. But, I'll try to illustrate the steps on how to solve these questions had there been Ka/Kb values.

For example, for HBr, the equilibrium in aqueous media is

HBr + H20 --> H30+ + Br-

Ka is defined as: [product]/[reactant] ([ ] means concentration)

Ka = [H30+] [Br-]/[HBr] (H20 is ignored)

The amount of moles of HBr is 5g/80g = 0.0625 mol HBr. In the equation we know the amount of HBr but not the amounts of H30+ (is the same as H+) nor Br-. In equilibrium, the amount of HBr decreases by a factor of x, while H30+ and Br- go from zero to x.

HBr + H20 --> H30+ + Br-
0.0625 -x x x

plug this in the Ka equation and solve for [H30+]

Ka = x . x/0.0625-x (If you have a Ka value, simply solve this equation for x. If the square root of the ratio of Ka/conc of reactant (i.e HBr) is less than 5%, round off 0.0625-x to 0.0625 only).

Once you find the H30+, simply find the pH by:

pH= -log[pH]

For NaOH, use the same steps with Kb and find [OH-]; but then, you find the pOH and subtract it from 14 to find the pH, since pH + pOH = 14

Hope this clears things up (and hope you'll give me 10 points!!)

Answer 5

Calculating the pH of these solutions will be easy because HBr is a strong acid, and NaOH is a strong base. First, you need to find the molarity of each solution. To do this, first find the number of moles you have.

moles= mass of sample/molecular weight

Then divide by volume (Convert to liters)

For part a, because HBr is a stong acid, it will completely dissociate in water. So [HBr]=[H]

Once you have the concentration, you are ready to find the pH

pH= -log[H]

For part b, the easiest thing to do would be to find the pOH

pOH= -log[OH]

pH=14-pOH

Answer 6

In either case you have complete dissociation, so the pH for a) and the pOH for b) will be the negative decadic logarithm of the educt concentration.

In order to get the concentration (i.e. mol/l) from the mass, you need to divide the mass of the educt (in grams) by its molar mass (in grams/mol), and then divide the result by the volume in L (which is 0.1L for a and 0.05L for b).

For case B, the molar mass of NaOH is 40 (23+16+1), so the concentration in mol/L calculates as
1.5 / 40 / 0.05 = 0.75

0.75 mol / L correspond to a pOH of roughly 0.2, which substracted from 14 yields a pH of roughly 13.8 (use a calculator to get the exact negative decadic logarithm of 0.75).

Answer 7

first you have to figure out the hydrogen ion , hydroxide ion concentration

to make this work you have to assume complete disassociation

the pH is defined as the inverse log of the hydrogen ion concentration

in the case of the NaOH solution, you have to use the relationship for hydroxide (OH) to (H) first, get the hydrogen ion concentration , and do the inverse log thing

good luck

Answer 8

just figure out how many grams in a ML. I dont have conversion chart in front of me, but when You figure this out, its simple.

Answer 9

That isnt a question - youll get no answers from that sunshine!