# working out the pH questions

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hey i was wondering if anyone knew the calculations to work out the approximate pH of:

NaHCO3

1 M HCl

10 M HCl

0.001 M HCl

and also how to write the equations to determine to pH of NaHCO3 and NaCH3C00?

Thankyou

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For the HCl solutions, HCl is a strong acid so whatever the concentration of the HCl is will be the concentration of the H+ ions in the solution. Therefore the pH would be the negative logarithm of the H+ concentration.

For the last two questions, you need to look at them as an equillibrium since they are both weak acids/bases. For a weak acid, you have to realize that it's conjugate base (acetate in this case) will want to grab a hydrogen atom from water and form the main acid again. As a result, any solution of the negative ion portion of a weak acid will be basic. Therefore, the equation you need to look at is that of the dissolution of a base. You will also need to look up the Kb of the acetate ion as the Kb is the equillibrium constant of the acetate ion reforming acetic acid. (If you don't have the Kb and you only have the Ka of acetic acid, remember that Ka*Kb = Kw = 1.0E-14). So when you write out your equation, you would write it like this: CH3COO- -> CH3COOH + OH- (This is not a fully balanced equation, but when dealing with acids/bases you just need to write the species which are involved in the reaction. We're not including water on the left hand side simply because there is so much water in the solution that the concentration really doesn't change).

So now we can write out our equillibrium expresssion; Kb(acetate) = ([CH3COOH][OH-])/[CH3COO-]

Initially, we have no acetic acid and no OH- in solution so their concentrations would be 0. As an example, we'll assume that we have a 1 molar solution of sodium acetate which would give us one mole of acetate ions. If we use an I.C.E. chart (Initial concentration, Change at Equillibrium, and Equillibrium concentration), the I portion would look like this; CH3COOH=0, OH-=0, CH3COO-=1. At equillibrium, the changes would be +x, +x, and -x as every mole of acetate forms one mole of OH- and CH3COOH. Finally, the equillibrium concentrations would be CH3COOH=x, OH-=x, CH3COO-=1-x. With the Kb of acetate being 5.6E-10, you just need to solve for x. Solving for x should give you a value of about 2.4E-5. That is the concentration of the OH- ions in solution. In order to figure out the pH, you need to know that pH+pOH = 14, and pOH = -Log([OH-]). With the numbers we've come up with, this gives a pOH value of about 4.6. 14-4.6 = pH = 9.4.

So the pH of a one molar solution of sodium acetate would be 9.4. The same process is used for any salt of a weak acid. If the salt is of a weak base, then you would use the Ka of the conjugate acid and adjust your equillibrium expression accordingly.

thanx

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