# What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? #K_a# acetic acid = #1.8 x 10^-5#?

Consequently,

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To determine the molar ratio of sodium acetate to acetic acid for preparing a buffer with pH = 4.5, you can use the Henderson-Hasselbalch equation:

[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

Given that the pH is 4.5 and the pKa of acetic acid is (1.8 \times 10^{-5}), we can rearrange the Henderson-Hasselbalch equation to solve for the ratio (\frac{[\text{A}^-]}{[\text{HA}]}):

[ 4.5 = -\log(1.8 \times 10^{-5}) + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ 4.5 = -(\log(1.8) + \log(10^{-5})) + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ 4.5 = -(0.2553 - 5) + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ 4.5 = 4.7447 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ 4.5 - 4.7447 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ -0.2447 = \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) ]

[ 10^{-0.2447} = \frac{[\text{A}^-]}{[\text{HA}]} ]

[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.2447} ]

[ \frac{[\text{A}^-]}{[\text{HA}]} ≈ 0.573 ]

Therefore, the molar ratio of sodium acetate to acetic acid should be approximately (0.573 : 1).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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