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Which of the following is the reasonable pH for 0.15 M 0f HCl?
a.) 0.15
b.)0.82
c.)8.24
d.)13.18
Which of the following is the reasonable pH for 0.15 M 0f NaOH?
a.) 0.15
b.)0.82
c.)8.24
d.)13.18
Answers
1) We know that for acidic solution [ H+ ] = √(KaC) = √(9.6 × 10-6 × 0.15) = √(1.44 × 10-6) = 0.0012M pH = - log [ H+ ] = - log [ 0.0012 ] = 2.920818
since NaOH is a strong electrolyte ,it will dissociate completely as follows -
2) NaOH == Na+ + OH-
conc. 0.5M 0.5M 0.5M
OH- conc. of NaOH is 0.5 , from this you can calculate pOH. Now you apply pOH+pH =14 (At 25 degree centigrade) to calculate pH of 0.5 M NaOH.
So, pOH=-log[OH-] =- log(0.5) =0.3010 Hence, pH=14 - pH =13.69 .
Answer:
Answer:
1) We know that for acidic solution [ H+ ] = √(KaC) = √(9.6 × 10-6 × 0.15) = √(1.44 × 10-6) = 0.0012M pH = - log [ H+ ] = - log [ 0.0012 ] = 2.920818
since NaOH is a strong electrolyte ,it will dissociate completely as follows -
2) NaOH == Na+ + OH-
conc. 0.5M 0.5M 0.5M
OH- conc. of NaOH is 0.5 , from this you can calculate pOH. Now you apply pOH+pH =14 (At 25 degree centigrade) to calculate pH of 0.5 M NaOH.
So, pOH=-log[OH-] =- log(0.5) =0.3010 Hence, pH=14 - pH =13.69 .
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