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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.301

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$= 2 \times M_{3} = \frac{2}{200} = \frac{1}{100} p H = - l o g_{10} (\frac{1}{100}) = - [l o g_{10} (1) - l o g_{10} (100)] p H = - [0 - 2] = 2$

Suggest Corrections

Similar questions

10 m L

\frac{M}{100} H_{2} S O_{4}

solution is mixed with

20 m L

\frac{M}{10} H_{2} S O_{4}

solution. Find the pH of resulting solution.

Q. 100 mL of a 0.2 M

H_{2} S O_{4}

solution, 200 mL of a 0.1 M

H N O_{3}

solution and 300 mL of a 0.2 M NaOH solution are mixed. The resultant solution will be

Q. 300 ml 0.1 M HCl and 200 ml of 0.03 M

H_{2} S O_{4}

are mixed. Calculate the normality of the resulting mixture.

Q. 200 ml of 1 M

H_{2} S O_{4}

, 300 ml of 3 M

H C l

and 100 ml of 2 M

H C l

are mixed and made up to 1 litre. The proton concentration in the resulting solution is:

20 m L

0.1 M

H_{2} S O_{4}

solution is mixed with

30 m L

0.2 M

N H_{4} O H

solution. The

p H

of the resultant solution is:
[

p K_{b}

N H_{4} O H = 4.7

]

Join BYJU'S Learning Program

Explore more

pH the Power of H

Standard XII Chemistry

Join BYJU'S Learning Program

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: