#3.00 * 10^-3# mol ofHBr are dissolved in water to make 16 L of solution. What is the concentration of hydroxide ions, [#OH^-#] in this solution?

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Answer 1 · 2017-07-16T00:23:58

#[HO^-]=5.33xx10^-11*mol*L^-1#

We know that #10^-14=[HO^-][H_3O^+]# under standard conditions at #298*K#......... And should we take #log_10# of BOTH SIDES, we gets......

#underbrace(log_10(10^-14))_-14=log_10[H_3O^+] + log_10[HO^-]#

And on rearrangement, #14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)#

And so our defining relationship, #pH+pOH=14#

Now.............

#[H_3O^+]=(3.00xx10^-3*mol)/(16*L)=1.875xx10^-4*mol*L^-1#

#pH=-log_10(1.875xx10^-4)=-(-3.727)=3.727#

And thus #pOH=10.27#, and......................

#[HO^-]=10^(-10.27)=5.33xx10^-11*mol*L^-1#