if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?

Chemistry

1 answer:

ipn [44]3 years ago

6 0

Answer:

M of HI = 5.4 M.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HI = (XMV) Ca(OH)₂.

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = 5.4 M.

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