If a 50.00 mL of 0.0155 M HI(aq) is mixed with 75.00 mL of 0.0106 M KOH(aq), the pH of the final solution is mathematically given as
pH=13.16
What is the pH of the final solution?
Generally, the equation for the Chemical reaction is mathematically given as
2HI+Ba(OH_2)--->BaI2+2H2O
Therefore
pOH=-log{OH}
pOH=-log(6.52)
pOH=0.8142
In conclusion, for the pH
pH+pOH=14
pH=14-0.8142
pH=13.16
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