10 mL of M200 H2SO4 is mixed with 40 mL of M200 H2SO4- The pH of resulting solution is-

Applying M_1V_1+M_2V_2=M_3V_3 (1200×101000)+ (1200×401000)=M_3×50/1000 M_3=1200 Now H_2SO_4⇌ 2H^++SO_3^2 Since 2H^+ are there so concentration =2× M_3=2200=1 ...

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Standard XIII

Chemistry

pH the Power of H

10 mL of M200...

Question

$10 mL$ of $\frac{M}{200} H_{2} S O_{4}$ is mixed with $40 mL$ of $\frac{M}{200} H_{2} S O_{4}$ . The pH of resulting solution is:

2.301

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1.301

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Solution

The correct option is C $2$
Applying
$M_{1} V_{1} + M_{2} V_{2} = M_{3} V_{3} (\frac{1}{200} \times \frac{10}{1000}) + (\frac{1}{200} \times \frac{40}{1000}) = M_{3} \times \frac{50}{1000} M_{3} = \frac{1}{200}$
Now $H_{2} S O_{4} ⇌ 2 H^{+} + S O_{3}^{2 -}$
Since $2 H^{+}$ are there so concentration
$= 2 \times M_{3} = \frac{2}{200} = \frac{1}{100} p H = - l o g_{10} (\frac{1}{100}) = - [l o g_{10} (1) - l o g_{10} (100)] p H = - [0 - 2] = 2$

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pH the Power of H

Standard XIII Chemistry

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10 mL of M200 H2SO4 is mixed with 40mL of M200 H2SO4. The pH of resulting solution is:

The correct option is C 2Applying M1V1+M2V2=M3V3(1200×101000)+(1200×401000)=M3×501000M3=1200 Now H2SO4⇌2H++SO2−3 Since 2H+ are there so concentration =2×M3=2200=1100pH=−log10(1100) =−[log10(1)−log10(100)]pH=−[0−2]=2

$10 mL$ of $\frac{M}{200} H_{2} S O_{4}$ is mixed with $40 mL$ of $\frac{M}{200} H_{2} S O_{4}$ . The pH of resulting solution is: