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`        Calculate the pH of a mixture containing 50ml of 1(N) HCl and 30 ml 1(N) NaOH .((Ans.. 0.6)). `
one year ago

Rohan Gujar
20 Points
```							Just use formula [H+]=(M1V1+M2V2)/(V1+V2) So, [H+]=(50-30)/80.    [N=M×n-factor][H+]=0.25.      So pH= -log[H+]                                   = -log(0.25)=0.6 Ans.
```
one year ago
Arun
24497 Points
```							Dear student Mole = concentration x volumemole of NaOH = 30ml x 1 = 30.0mole of HCl = 50ml x 1= 50.030 mole of NaOH exactly neutralized by 30 mole of HCl.NaOH + HCl = NaCl + H2Othat is 50 -30 = 20 HCl moleCon of H+ = no of mole of H+ / total volume of solution (50+ 30 ml = 80 ml)= 20/80 = 0.25pH = - log ( H+)= - log (1/4) = 0.6021s0 the pH of solution will be 0.6021  RegardsArun
```
one year ago
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