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MolSynapse / Question bank / Equivalent Concept and Titration / Acid-base titrations and double titration (NaOH, Na2CO3, NaHCO3)

JEE Main levelSingle correct optionDifficulty 2 of 3eq129

50.0 mL of 0.200 M HCl is mixed with 30.0 mL of 0.200 M NaOH. Assuming volumes are additive, the pH of the resulting solution is (take $lo g 2 = 0.301$ ):

A $1.00$
B $1.60$
C $1.30$
D $2.00$

Answer: C

Moles of

H X^{+}

= 10.0 mmol and of

OH X^{-}

= 6.0 mmol, so 4.0 mmol

H X^{+}

is left in 80.0 mL.

[H X^{+}] = 4.0/80.0 = 0.0500

M, and pH =

- lo g (0.0500) = 2 - lo g 5 = 1.30

.

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