MMLU-Pro Question 4131

Okay, let's tackle this pH problem for 0.15 M H₂SO��. Hmm, sulfuric acid is a diprotic acid, right? So it donates two protons. The first proton is strong, meaning it completely dissociates. The second proton has a Ka2 of 1.26 × 10^-2. Wait, but the problem says K₂ = 1.26 × 10^-2. So K₂ is Ka2. First, I remember that for H₂SO��, the first dissociation is complete: H₂SO�� → H⁺ + HSO��⁻. So that gives 0.15 M H⁺ from the first step. Then the HSO��⁻ can further dissociate as HSO��⁻ �� H⁺ + SO��²⁻, with Ka2 = 1.26 × 10^-2. So the total H⁺ concentration will be 0.15 M from the first dissociation plus whatever comes from the second step. But wait, the second dissociation is partial. However, because the first dissociation gives a significant amount of H⁺, the second one might be suppressed. But Ka2 is 1.26e-2, which is not too small. So maybe we can't ignore the second dissociation. Let me think. Let's denote the initial concentration after the first dissociation: [H⁺] = 0.15 M, [HSO��⁻] = 0.15 M. Then when HSO��⁻ dissociates, it will produce x more H⁺ and x SO��²⁻. So the equilibrium concentrations would be: [H⁺] = 0.15 + x [HSO��⁻] = 0.15 - x [SO��²⁻] = x Ka2 = [H⁺][SO��²⁻] / [HSO��⁻] = (0.15 + x)(x) / (0.15 - x) = 1.26 × 10^-2 Hmm, since Ka2 is 0.0126, which is not too small compared to 0.15. So x might be significant. Let's see. If we assume x is small, then (0.15)(x)/0.15 = x = 0.0126. But x = 0.0126 is not negligible compared to 0.15, because 0.0126 / 0.15 �� 0.084, which is 8.4%, so maybe the approximation isn't great. So better to solve the quadratic equation. So set up the equation: (0.15 + x)(x) / (0.15 - x) = 0.0126 Multiply both sides by (0.15 - x): (0.15 + x)(x) = 0.0126(0.15 - x) Expand left side: 0.15x + x² Right side: 0.00189 - 0.0126x Bring all terms to left: 0.15x + x² - 0.00189 + 0.0126x = 0 Combine like terms: x² + 0.1626x - 0.00189 = 0 Now, quadratic equation: x = [-b ± sqrt(b² - 4ac)]/(2a) Here a=1, b=0.1626, c=-0.00189 Discriminant: (0.1626)^2 + 4*0.00189 Calculate 0.1626^2: approx 0.0264 4*0.00189 = 0.00756 Total discriminant �� 0.0264 + 0.00756 = 0.03396 sqrt(0.03396) �� 0.184 So x = [-0.1626 ± 0.184]/2 Take the positive root: (-0.1626 + 0.184)/2 = 0.0214 / 2 �� 0.0107 So x �� 0.0107 M Therefore total [H⁺] = 0.15 + 0.0107 �� 0.1607 M Then pH = -log(0.1607) �� ? Let me compute log(0.1607). Since log(0.1) = -1, log(0.16) �� log(1.6e-1) = log(1.6) -1 �� 0.204 -1 = -0.796 But 0.1607 is a bit more than 0.16. Let's see: log(0.16) = -0.7959 log(0.1607) �� ? Difference between 0.16 and 0.1607 is 0.0007. The derivative of log(x) is 1/(x ln10). So approximate delta log �� (0.0007)/(0.16 * 2.3026) �� 0.0007 / (0.3684) �� 0.0019. So log(0.1607) �� -0.7959 + 0.0019 �� -0.794. Wait, but since x increases, log increases. Wait, 0.1607 is larger than 0.16, so log(0.1607) is less negative. So log(0.16) = -0.7959, then log(0.1607) �� -0.7959 + (0.0007 / 0.16) * (1 / ln(10)) ? Wait, maybe better to compute 10^-0.8 = approx 0.1585. Hmm, 10^-0.8 �� 0.1585. So 0.1607 is slightly higher. Let's see: Let x = 0.1607 log(x) = ? We know that log(0.1585) = -0.8 The difference between 0.1607 and 0.1585 is 0.0022. Approximate the derivative of log(x) at x=0.1585: d(log(x))/dx = 1/(x ln10) �� 1/(0.1585 * 2.3026) �� 1/(0.365) �� 2.74 So delta log �� 2.74 * 0.0022 �� 0.0060 So log(0.1607) �� -0.8 + 0.006 = -0.794 Therefore, pH = -log(0.1607) �� 0.794 So approximately pH �� 0.79 Looking at the options, option B is 0.79. So the answer should be B. Wait, but let me check again. Wait, the problem says "assuming K2 =1.26 × 10^-2". So my calculation leads to pH �� 0.79, which is option B. But wait, let me verify the calculation again. Wait, if total H+ is 0.1607, then pH is -log(0.1607) = ? Let me calculate it more accurately. Compute log(0.1607): We know that log(0.16) = log(16/100) = log(16) - 2 = 1.2041 - 2 = -0.7959 Wait, log(16) is log(2^4) = 4*log(2) �� 4*0.3010 = 1.204, yes. So 0.16 = 1.6e-1, so log(0.16)=log(1.6)-1≈0.2041-1= -0.7959 Now, 0.1607 is 0.16 * (1 + 0.007/0.16) = 0.16*(1 + 0.04375) log(0.1607)=log(0.16) + log(1.04375) log(1.04375)≈0.0185 (using log(1+x)≈x/(ln(10)) for small x? Wait, actually log10(1+x)≈x/(2.3026) when x is small. Wait, x=0.04375. Wait, log10(1+x) = ln(1+x)/ln(10) �� (x - x²/2)/ln(10) But maybe approximate: log10(1.04) �� 0.0170 log10(1.04375) �� ? Let's use linear approximation between 1.04 and 1.05. log(1.04)=0.0170 log(1.05)=0.0212 Difference per 0.01 is 0.0042 So 0.04375 - 0.04 = 0.00375, so 0.00375 / 0.01 = 0.375 of the interval. So 0.0170 + 0.375*(0.0042) �� 0.0170 + 0.001575 = 0.018575 So log(1.04375)≈0.0186 Therefore, log(0.1607)= -0.7959 + 0.0186 = -0.7773? Wait, that doesn't make sense. Wait, no: Wait, 0.1607 is 0.16 * 1.004375? Wait, no: Wait, 0.1607 / 0.16 = 1.004375. Wait, no, 0.16 * 1.004375 = 0.1607. Wait, but 0.16 * 1.004375 = 0.16 + 0.16*0.004375 = 0.16 + 0.0007 = 0.1607. So x=0.0007. Wait, but earlier I thought 0.16 to 0.1607 is 0.0007 increase. But that would be 0.16 * (1 + 0.004375). Wait, perhaps I made a mistake here. Let's do it properly. Let me compute 0.1607. log(0.1607) = log(1.607 × 10^-1) = log(1.607) + log(10^-1) = log(1.607) - 1 Compute log(1.607): We know that log(1.6) = 0.2041 log(1.607) = log(1.6 + 0.007) �� log(1.6) + (0.007 / (1.6 * ln(10))) Wait, derivative of log(x) at x=1.6 is 1/(x ln(10)). So delta log �� delta x / (x ln(10)) delta x = 0.007, x=1.6 ln(10)=2.3026 so delta log �� 0.007 / (1.6 * 2.3026) �� 0.007 / 3.684 �� 0.0019 So log(1.607) �� 0.2041 + 0.0019 = 0.206 Thus log(0.1607)=0.206 -1 = -0.794 Therefore, pH = -(-0.794) = 0.794, which rounds to 0.79. So pH≈0.79. So option B. But wait, another way: let's calculate -log(0.1607) Using calculator-like approach: We know that 10^-0.79 = ? 10^-0.8 = 0.1585 10^-0.79 = 10^(-0.8 + 0.01) = 10^-0.8 * 10^0.01 �� 0.1585 * 1.0233 �� 0.162 Wait, 10^0.01 �� 1.0233 So 0.1585 * 1.0233 �� 0.1585 + 0.1585*0.0233 �� 0.1585 + 0.0037 �� 0.1622 Which is very close to our 0.1607. So 0.1607 is slightly less than 0.1622, which corresponds to pH slightly higher than 0.79. Let's see: Suppose 10^-0.79 = 0.1622 We have 0.1607, which is lower, so pH is higher than 0.79. Compute the difference: 0.1622 - 0.1607 = 0.0015 Using derivative: d(pH)/d[H+] = -1/( [H+] ln(10) ) But maybe better to do: Let x = 0.1607 We want to find p = -log(x) We know that at p=0.79, x=0.1622 The difference in x is -0.0015 The derivative of x with respect to p is dx/dp = -x ln(10) So dp = dx / (-x ln(10)) = -dx/(x ln(10)) dx = 0.1607 - 0.1622 = -0.0015 dp = -(-0.0015)/(0.1622 * 2.3026) �� 0.0015 / (0.373) �� 0.004 So p �� 0.79 + 0.004 �� 0.794, so pH≈0.794, which rounds to 0.79. So yes, option B. Therefore, the answer is B.

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