Calculating pH from Concentration of Strong Acid/Base

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Michelle N - 2C
Posts: 117
Joined: Wed Sep 18, 2019 12:19 am

Calculating pH from Concentration of Strong Acid/Base

Postby Michelle N - 2C » Tue Jan 21, 2020 12:03 am

On our lecture on January 13th, Lavelle stated on his notes that “If we know the concentration of the strong base/acid, we can calculate pH or pOH.” He then proceeded to the example that 0.0030 M of Ba(OH)2 gives 0.0060 M of OH-. How did that happen? Can someone explain to me this specific process? Thank you!

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: Calculating pH from Concentration of Strong Acid/Base

Postby nicolely2F » Tue Jan 21, 2020 12:08 am

Since Ba(OH)2 is a strong acid, it is fully dissociated in water. This means that, at equilibrium, zero Ba(OH)2 is left and there is only Ba2+ and OH- in solution, dissociated from the quantity of reactant added to the water.
If you write down the equation for this reaction, you will notice that the coefficients are 1, 1, 2 for Ba(OH)2, Ba2+, OH-. Coupling this with the fact that Ba(OH)2 dissociates 100%, we know that we will have twice as much of OH- as we did of Ba(OH)2. In other words, if we add X mol of Ba(OH)2 to water, the result will be of X mol of Ba2+ and 2X mol of OH-.

WesleyWu_1C
Posts: 117
Joined: Thu Jul 25, 2019 12:16 am

Re: Calculating pH from Concentration of Strong Acid/Base

Postby WesleyWu_1C » Tue Jan 21, 2020 12:09 am

When Ba(OH)2 is placed in water, it dissociates into Ba^2+ and 2OH-. So the reaction would be Ba(OH)2 + H2O -> H2O + Ba^2+ + 2OH- For every 1 mole of Ba(OH)2, there are 2 moles of OH-.


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