When a volume of 60.0 mL of 0.200 M HBr is mixed with a volume of 30.0 mL of 0.400 M CH3NH2, The pH value is mathematically given as
pH=10.64
What is the pH value when 60.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂?
Question Parameters:
The pH when 60.0 mL of 0.200 M HBr
30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10^{-4}).
Generally, the equation for the Chemical Reaction is mathematically given as
H Br + H_3C NH_2----- > CH_3 NH_3 Br
Therefore
oH=p^{kb}+-log[tex]\frac{salt}{base}[/tex]
OH=-log(4.4*10^{-4})+[tex]\frac{0.1}{0.1}[/tex]
OH=3.36
In conclusion, The equation pH value
pH+OH=14
Therefore
pH+=14-3.36
pH=10.64
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