Question

#try this..... qsn. ​

Answer 1

conc of acid is more.

$[H^+] = \frac{ M_1V_1 - M_2 V_2}{V_1 + V_2}\\\\= \frac{ 2*10^{-3} *1 - 5*10^{-4}*2}{ 2 + 1}\\\\= \frac{(2-1) * 10^{-3}}{3}\\\\\frac{10^-3}{3}$

pH = -log[H⁺] = 3.48

Answer 2

Answer:

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