10 mL of M100H2SO4 solution is mixed with 20 mL of M10 H2SO4 solution- Find the pH of resulting solution-

The final concentration (C_f) is given as: C_1V_1+C_2V_2=C_fV_f (1/100×10/1000)+(1/10×20/1000)=C_f×30/1000 C_f=0.07 M Now since, H_2SO_4⇌ 2H^++SO_4^2 2H^+ io ...

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Standard XIII

Chemistry

pH the Power of H

10 mL of M100...

Question

$10 m L$ of $\frac{M}{100} H_{2} S O_{4}$ solution is mixed with $20 m L$ of $\frac{M}{10} H_{2} S O_{4}$ solution. Find the pH of resulting solution.

0.85

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0.69

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Solution

The correct option is A 0.85
The final concentration ( $C_{f}$ ) is given as:
$C_{1} V_{1} + C_{2} V_{2} = C_{f} V_{f} (\frac{1}{100} \times \frac{10}{1000}) + (\frac{1}{10} \times \frac{20}{1000}) = C_{f} \times \frac{30}{1000} C_{f} = 0.07 M Now since, H_{2} S O_{4} ⇌ 2 H^{+} + S O_{4}^{2 -} 2 H^{+} ions are there . So concentration = 2 \times C_{f} = 2 \times 0.07 = 0.14 p H = - {log}_{10} (0.14) p H = 0.85$

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pH the Power of H

Standard XIII Chemistry

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10 mL of M100H2SO4 solution is mixed with 20 mL of M10 H2SO4 solution. Find the pH of resulting solution.

The correct option is A 0.85The final concentration (Cf) is given as: C1V1+C2V2=CfVf(1100×101000)+(110×201000)=Cf×301000Cf=0.07 MNow since, H2SO4⇌2H++SO2−42H+ ions are there.So concentration=2×Cf=2×0.07=0.14pH=−log10(0.14)pH=0.85

$10 m L$ of $\frac{M}{100} H_{2} S O_{4}$ solution is mixed with $20 m L$ of $\frac{M}{10} H_{2} S O_{4}$ solution. Find the pH of resulting solution.