Question #111140

Calculate the pH of the resulting solution if 20.0 mL of 0.200 M HCl(aq) is added to 30.0 mL of 0.200 M NaOH(aq).


Calculate the pH of the resulting solution if 20.0 mL of 0.200 M HCl(aq) is added to 10.0 mL of 0.300 M NaOH(aq).

Expert's answer

(I) effective molarity is given by -

MV=M1V1−M2V2MV=M_1V_1-M_2V_2

Where


M1=0.2,M2=0.2,V1=30,V2=20 mlM_1=0.2,M_2=0.2,V_1=30,V_2=20\ ml

Now


M(20+30)=2M=0.04M(20+30)= 2\\M=0.04

Molarity = OH−OH^- ion concentration = 0.04


pOH=log(1OH−)=log(25)=1.398pOH=log(\frac{1}{OH^-})=log(25)= 1.398

pH= 14-pOH= 14-1.398=12.602


(II)

Similarly

MV=M1V1−M2V2MV=M_1V_1-M_2V_2


M(20+10)=20×0.2−10×0.3M=0.033M(20+10)=20\times0.2-10\times0.3\\M=0.033

Here molarity = H+H^+ concentration = 0.033


pH=log(1H+)=log(30)=1.477pH= log(\frac{1}{H^+})= log(30)=1.477


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